{"id":2163,"date":"2023-05-01T06:57:55","date_gmt":"2023-05-01T06:57:55","guid":{"rendered":"https:\/\/vartmaaninstitutesirsa.com\/?page_id=2163"},"modified":"2024-06-29T00:23:05","modified_gmt":"2024-06-29T00:23:05","slug":"chapter-4-moving-charges-and-magnetism","status":"publish","type":"page","link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/chapter-4-moving-charges-and-magnetism\/","title":{"rendered":"Chapter 4 Moving Charges and Magnetism"},"content":{"rendered":"
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\n<\/div>\n

Chapter\u20134:<\/span><\/u><\/strong>\u00a0\u00a0<\/span><\/u><\/strong>Moving Charges and Magnetism<\/span><\/u><\/strong><\/h1>\n

1. Electromagnetism<\/span><\/u><\/strong><\/p>\n

The branch of physics which deals with the study of\u00a0<\/span><\/em>magnetism due to electricity<\/span><\/u><\/em>\u00a0is called electromagnetism<\/span><\/em>. The relationship between electricity & Magnetism was firstly given by Oersted which is as below<\/span><\/p>\n

 <\/p>\n

2. Oersted experiments:-<\/span><\/u><\/strong>\u00a0<\/span><\/p>\n

\"F:\\unitAs shown in fig, Oersted found a relation between current & magnetism, when a magnetic niddle is placed parallel to direction of flow of current then the niddle suffer a deflection,\u00a0<\/span><\/p>\n

When the direction of flow of current is reversed then the deflection is also reverses, the direction of deflection can be given by Ampere\u2019s swimming rule.<\/span><\/p>\n

 <\/p>\n

Ampere\u2019s swimming rule:-<\/span><\/u><\/strong><\/p>\n

\"F:\\unitAccording to this rule, if we assume that a man is swimming in direction of flow of current, such that current is flowing from his feet to head then the north pole of niddle will deflect towards his left hand.<\/span><\/p>\n

 <\/p>\n

3. Magnetic field:-<\/span><\/u><\/strong><\/p>\n

The space around a magnet or a current carrying conductor in which its magnetic effect can be experienced is called magnetic field<\/span><\/em>. It is a vector quantity & denoted by<\/span>\"\".<\/span><\/p>\n

Force on a moving charge in a magnetic field or Magnetic Lorentz force<\/span><\/u><\/strong>.<\/span><\/u>m.imp<\/sup><\/span><\/u><\/strong>\u00a0<\/span><\/u><\/p>\n

Suppose a +q charge moving in a magnetic field\u00a0<\/span>\"\"\u00a0with velocity<\/span>\"\". Then\u00a0<\/span>\"\"\u00a0is the magnetic force experienced by the charge depends on<\/span><\/p>\n

\"F:\\unit\u00a0F<\/span>m<\/sub><\/span>\u00a0\u00a0\u00a0\u00a0<\/sub><\/span>\"\"\u00a0<\/span><\/strong>q<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>(i)<\/span><\/p>\n

F<\/span>m\u00a0<\/sub><\/span>\"\"\u00a0<\/sub><\/span>\"\"\u00a0sin\u03b8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>(ii)<\/span><\/p>\n

F<\/span>m\u00a0<\/sub><\/span>\"\"\u00a0B<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>(iii)<\/span><\/p>\n

Combining eq<\/span>n<\/sup><\/span>s we get<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

F<\/span>m\u00a0<\/sub><\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sub><\/span>q<\/span>\"\"\u00a0B Sin\u03b8\u00a0<\/span><\/p>\n

Or<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>F<\/span>m<\/sub><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sub><\/span>=<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>K q\u00a0<\/span>\"\"\u00a0B Sin\u03b8\u00a0<\/span><\/p>\n

Here<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>K = 1<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

So Fm<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sub><\/span>=<\/sub><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sub><\/span>q<\/span>\"\"\u00a0B Sin\u03b8<\/span><\/p>\n

Or in vector form<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sub><\/span><\/strong>=<\/sub><\/span><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>q (<\/span><\/strong>\"\"\u00a0<\/span><\/strong>\"\")<\/span><\/strong>\u00a0<\/span><\/p>\n

The direction of\u00a0<\/span><\/em>\"\"\u00a0<\/sub><\/span><\/em>can<\/span><\/em>\u00a0<\/sub><\/span><\/em>be given by Right hand rule.<\/span><\/em>\u00a0\u00a0<\/span><\/em><\/p>\n

I,e.\u00a0<\/span><\/em>\"\"\u00a0\u00a0<\/sub><\/span><\/em>is perpendicular to plane containing both\u00a0<\/span><\/em>\"\"\u00a0&\u00a0<\/span><\/em>\"\".\u00a0<\/span><\/em><\/p>\n

Also direction of\u00a0<\/span><\/em>\"\"\u00a0\u00a0<\/sub><\/span><\/em>\u00a0<\/span><\/em>is given by\u00a0<\/span><\/em>Fleming left hand rule<\/span><\/em><\/strong>.<\/span><\/p>\n

 <\/p>\n

Definition of B:-<\/span><\/u><\/strong>\u00a0\u00a0<\/span><\/p>\n

If\u00a0<\/span>\"\"\u00a0=1,<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>q = 1<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>&<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

\"\"\u00a0\u00a0<\/span>Then<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

Hence magnetic field is equal to the force experienced by a unit charge moving with a unit velocity perpendicular to the direction of magnetic field at that point.<\/span><\/em><\/p>\n

 <\/p>\n

Special cases:-<\/span><\/strong><\/p>\n

(i)-<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If \u03b8 = 0<\/span>0<\/sup><\/span>\u00a0or 180<\/span>0\u00a0<\/sup><\/span>\u00a0<\/span>\u00a0<\/sup><\/span>then<\/span>\u00a0\u00a0\u00a0<\/span>sin0<\/span>0\u00a0<\/sup><\/span>=<\/span>\u00a0\u00a0<\/span>0<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\u21d2\u00a0\u00a0<\/span>\"\"0<\/span><\/p>\n

Hence a charge particle moving along or opposite to direction of magnetic field experience no any force.<\/span><\/em><\/p>\n

(ii)-<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

Means a rest charge in magnetic field experienced no any force.<\/span><\/em><\/p>\n

(iii)-<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>If<\/span>\u00a0\u00a0<\/span>\u03b8 = 90<\/span>0<\/sup><\/span>\u00a0\u00a0<\/span>\u21d2\u00a0\u00a0\u00a0<\/span>sin 90<\/span>0<\/sup><\/span>\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0\u00a0<\/span>max<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\uf0f0<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0maximum.<\/span><\/p>\n

Hence a charge particle moving perpendicular to external magnetic field experience maximum force.<\/span><\/em><\/p>\n

 <\/p>\n

Unit of magnetic field B:<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

SI unit of B is\u00a0<\/span>Tesla<\/span><\/u>\u00a0\u00a0\u00a0<\/span><\/u>or<\/span><\/u>\u00a0\u00a0\u00a0<\/span><\/u>Weber\/metre<\/span><\/u>2<\/sup><\/span><\/u>\u00a0<\/span><\/p>\n

As\u00a0<\/span>\u00a0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>B =\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

Or<\/span>\u00a0\u00a0\u00a0<\/span>1 Tesla =<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

Hence magnetic field or magnetic induction is said to be one Tesla if a charge of one 1 coulomb moving perpendicular to magnetic field with a velocity of 1ms<\/span><\/em>-1<\/sup><\/span><\/em>\u00a0experience a force of one Newton.<\/span><\/em><\/p>\n

 <\/p>\n

Dimensional formula<\/span><\/strong>: –<\/span>\u00a0\u00a0<\/span><\/p>\n

B =<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

=<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

= [MA<\/span>-1<\/sup><\/span>\u00a0T<\/span>-2<\/sup><\/span>]<\/span><\/p>\n

In c,g,s system the unit of magnetic field is gauss G.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

1 T =<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>10<\/span><\/strong>4<\/sup><\/span><\/strong>G<\/span><\/strong><\/p>\n

\u00a0<\/span><\/strong><\/p>\n

4. Right hand rule or right hand screw rules:-<\/span><\/u><\/strong>\u00a0<\/sup><\/span><\/u><\/strong><\/p>\n

According to right hand rule if we curl the finger of our right hand from\u00a0<\/span>\"\"\u00a0to\u00a0<\/span>\"\"\u00a0then thumb will give the direction of\u00a0<\/span>\"\"m<\/sub><\/span><\/strong>.<\/sub><\/span><\/p>\n

 <\/p>\n

\"Physics\"5 Fleming\u2019s left hand rule:-<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong>\u00a0<\/span><\/u><\/strong><\/p>\n

If we stretch the first finger, the central finger & thumb of left hand mutually perpendicular to each other, such that first finger points to the direction of magnetic field, the central finger in direction to the electric current then thumbs will give the direction of force experienced by the charge particle.<\/span><\/p>\n

 <\/p>\n

6 Biot Savart\u2019s Law:-<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong>\u00a0<\/span><\/u><\/strong><\/p>\n

As we know when current passes through a conductor, then magnetic field set up around the conductor which was firstly calculated by Jean Biot & Felix Savart called Biot Savart\u2019s Law.<\/span><\/p>\n

\"F:\\unitAccording to this law, the magnitude of magnetic field d<\/span>\"\"\u00a0at a point due to current following through a conductor is depends upon, current , length of conductor, distance of point from conductor & angle between point and conductor as<\/span><\/p>\n

\u00a0\u00a0<\/span>\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Combining all eq<\/span>n<\/sup><\/span>s we get<\/span>\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

\u00a0<\/span>\"\"<\/p>\n

Or<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

Where<\/span>\u00a0\u00a0<\/span>K<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"\u00a0= 10<\/span>-7<\/sup><\/span>\u00a0\u00a0<\/sup><\/span>TmA<\/span>-1<\/sup><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sup><\/span><\/p>\n

\u00a0<\/sup><\/span>Here u<\/span>0<\/sub><\/span>\u00a0called\u00a0<\/span>permeability of free space<\/span><\/u><\/em>.\u00a0<\/span><\/p>\n

\"\"So Biot Savart\u2019s law become\u00a0<\/span><\/p>\n

\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

In vector form<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span><\/strong>=<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>\"\"<\/p>\n

The direction of dB can be given by right hand rule.\u00a0<\/span><\/p>\n

Imp.<\/sup><\/span><\/em><\/strong>\u00a0<\/sup><\/span><\/strong>In present case the direction of\u00a0<\/span><\/em>\"\"\u00a0is perpendicular to plane of paper containing both\u00a0<\/span><\/em>\"\"\u00a0&\u00a0<\/span><\/em>\"\"\u00a0in inward direction; if P point lies in L.H.S then in outwards direction.<\/span><\/em><\/p>\n

Special cases:-<\/span><\/u><\/strong><\/p>\n

    \n
  1. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u00a0 If\u00a0\u00a0\u00a0 \"\"<\/li>\n<\/ol>\n

    \"\"<\/p>\n

      \n
    1. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u00a0\u00a0<\/span>If\u00a0<\/span>\"\"<\/li>\n<\/ol>\n

      \"\"<\/p>\n

      \u00a0\u00a0<\/span>I,e.\u00a0<\/span>magnetic field at a point perpendicular to current element is maximum<\/span><\/em><\/p>\n

      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>This law is applicable for infinite small conductor.<\/span><\/p>\n

      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>This law is similar to coulomb\u2019s law in electrostatics.<\/span><\/p>\n

      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>This law is difficult to verify experimentally because it is difficult to construct infinite small conductor.<\/span><\/p>\n

       <\/p>\n

      7 Biot Savart\u2019s law V\/s Coulomb\u2019s Law:-<\/span><\/u><\/strong><\/p>\n

      As according to Coulomb\u2019s Law<\/span>\u00a0\u00a0<\/span><\/p>\n

      \u00a0\u00a0<\/span>\"\"=<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

      & from Biot Savart\u2019s law<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

      dB =<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

      On comparing above eq<\/span>n<\/sup><\/span>s we can give similarity & dissimilarity between both.<\/span><\/p>\n

      Similarity<\/span><\/u><\/p>\n

        \n
      1. Both the law obey inverse square law (\u03b1 \"\")<\/li>\n
      2. Both the law obeys principle of Superposition.<\/li>\n<\/ol>\n

        Dissimilarity:-<\/span><\/u><\/p>\n

          \n
        1. The source of magnetic field is current while source of electric field is charged.<\/li>\n
        2. Electric field exists due to both rest & motion of charge while magnetic field exists due to motion of charge.<\/li>\n
        3. The magnetic field is angle dependent while electric field is not.<\/li>\n
        4. <\/li>\n<\/ol>\n

          8 Magnetic field due to a long straight Conductor carrying current:-<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong><\/p>\n

          Suppose a point P having a distance a from the straight conductor carrying current I, then according to Biot savart Law, the magnetic field\u00a0<\/span>\"\"\u00a0\u00a0<\/span>at P due to small current element\u00a0<\/span>\"\",<\/span>\u00a0\u00a0\u00a0<\/span>r distance apart is<\/span><\/p>\n

          \"F:\\unit\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>–(1)<\/span><\/p>\n

          Now from \u0394 COP<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

          \"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

          Or<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>(i)<\/span><\/p>\n

          Also<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

          \"\"\u00a0\u00a0\u00a0\u00a0<\/span>(ii)<\/span><\/p>\n

          As<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>tan<\/span>\"\"\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"<\/p>\n

          \u21d2\u00a0\u00a0\u00a0<\/span>\"\"\u00a0= a tan<\/span>\"\"<\/p>\n

          Differentiating both side d<\/span>\"\"\u00a0=a sec<\/span>2<\/sup><\/span>\u00a0<\/span>\"\".d<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>(iii)<\/span><\/p>\n

          Using all values in eq<\/span>n<\/sup><\/span>s (1) we get<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

          \u00a0<\/span>\"\"=<\/span>\"\"<\/p>\n

          Or<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

          To calculate complete magnetic field integrating both sides we get<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

          B =\u00a0<\/span>\"\"\u00a0<\/span>\"\"<\/p>\n

          B =\u00a0<\/span>\"\"\u00a0<\/span>\"\"<\/p>\n

          B =<\/span>\"\"\u00a0<\/span>\"\"\u00a0[Sin<\/span>\"\"2\u00a0<\/sub><\/span>\u2013 Sin(-<\/span>\u2205<\/span>1<\/sub><\/span>)]<\/span><\/p>\n

          Or<\/span>\u00a0\u00a0\u00a0<\/span>B =<\/span><\/strong>\"\"\u00a0<\/span><\/strong>\"\"\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>[Sin<\/span><\/strong>\u2205<\/span><\/strong>1\u00a0<\/sub><\/span><\/strong>+ Sin<\/span><\/strong>\u2205<\/span><\/strong>2<\/sub><\/span><\/strong>]<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/p>\n

          The direction of B can be given by\u00a0<\/span>Right hand rule<\/span><\/em>.<\/span><\/p>\n

          Special cases:-<\/span><\/u><\/p>\n

            \n
          1. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If the conductor is infinite long & point P lies near to the dipole then<\/li>\n<\/ol>\n

            B =<\/span>\"\"\u00a0<\/span>\"\"\u00a0[sin90<\/span>0<\/sup><\/span>\u00a0<\/sub><\/span>+ sin90<\/span>0<\/sup><\/span>]<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

            Or<\/span>\u00a0<\/span>B =<\/span><\/strong>\"\"\u00a0<\/span><\/strong>\"\"=\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/p>\n

              \n
            1. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span> If the conductor is infinite long but P lies near end y then\u00a0 \u2205<\/span>1<\/sub><\/span>=\u00a0 90,\u00a0 \u2205<\/span>2\u00a0<\/sub><\/span>=0<\/li>\n<\/ol>\n

              B =\u00a0<\/span>\"\"\u00a0[sin90<\/span>\u00a0<\/sub><\/span>+ sin0]<\/span>\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              \u00a0<\/span>=<\/span>\"\"\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

              \"\"Q1. A current of 10A is flowing from east to west in a long straight wire kept on a horizontal table. The magnetic field developed at a distance 10cm vertically above the wire is:\u00a0<\/span><\/strong><\/p>\n

              Ans:\u00a0<\/span>\"\"\u00a0\u00a0<\/span><\/p>\n

              \u00a0<\/span>So\u00a0<\/span>\"\"<\/p>\n

              Q2. A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B (Fig. 4.3). What is the magnitude of the magnetic field?<\/span><\/strong><\/p>\n

              Solution we find that there is an upward force F, of magnitude<\/span>\"\", for mid-air suspension, this must be balanced by the force due to gravity:<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span><\/p>\n

              \u21d2<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

              \"\"0.65 T<\/span><\/p>\n

              Note that it would have been sufficient to specify<\/span>\"\", the mass per unit length of the wire. The earth\u2019s magnetic field is approximately 4 \u00d7 10<\/span>\u20135<\/sup><\/span>T and we have ignored it.<\/span><\/p>\n

              \"\"\u00a0Q3.<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>An element<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/strong>is placed at the origin and carries a large current I = 10 A. What is the magnetic field on the y-axis at a distance of 0.5 m. \u2206x = 1 cm.\u00a0<\/span><\/strong><\/p>\n

              Solution<\/span>\"\"\u00a0\u00a0<\/span>here\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

              \"\"<\/p>\n

              \"\"\u00a0,<\/span><\/p>\n

              \u00a0<\/span>\"\"<\/p>\n

              \u21d2<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

              \"\"\u00a0\u00a0<\/span>\"\"<\/p>\n

              The direction of the field is in the +z-direction.\u00a0<\/span><\/p>\n

              As<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

              \"\"<\/p>\n

               <\/p>\n

              9 Magnetic field At the Center Of The Circular Coil Carrying current:-<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong><\/p>\n

              \"F:\\unitAccording to Biot Savart Law, the magnetic field at the center of circular coil carrying current can be given as<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

              \u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>–(1)<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              \u00a0<\/span>(Here\u00a0<\/span>\"\"\u00a0is angle between current element Id<\/span>\"\"\u00a0& r. )<\/span><\/p>\n

              As radius is always perpendicular to tangent\u00a0<\/span><\/p>\n

              so<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u03b8= 90<\/span>0<\/sup><\/span>\u00a0<\/span>\u21d2<\/span>\u00a0 \u00a0<\/span>sin90= 1<\/span><\/p>\n

              So<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

              To calculate complete magnetic field, integrate both sides\u00a0<\/span><\/p>\n

              \u21d2\u00a0\u00a0\u00a0\u00a0 B =<\/span>\"\"\u0283<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              Here \u0283<\/span>\"\"\u00a0circumference of circle<\/span>\u00a0\u00a0<\/span><\/p>\n

              \u21d2<\/span> \u00a0\u00a0 B =<\/span>\"\"r<\/span><\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>B =<\/span>\"\"<\/p>\n

              If the circular coil contains n turns then<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              B =\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              or<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>B=\u00a0<\/span><\/strong>\"\"<\/p>\n

              Here the direction of magnetic field can be given by right hand rule.<\/span>\u00a0<\/span><\/p>\n

              Q4. The magnetic dipole moment of a current carrying coil does not depend upon:<\/span><\/strong><\/p>\n

              A Number of turns of the coil<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              B Cross-sectional area of the coil<\/span><\/p>\n

              C Current flowing in the coil<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              D Material of the turns of the coil<\/span><\/strong><\/p>\n

              Q5. An electron is revolving around the nucleus in a circular orbit with a speed of\u00a0<\/span><\/strong>\"\"\u00a0m\/s . If the radius of the orbit is\u00a0<\/span><\/strong>\"\"\u00a0m, find the current constituted by the revolving electron in the orbit<\/span><\/strong><\/p>\n

              Solution\u00a0<\/span>\"\"<\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              Q6. A steady current of 2A flows through a circular coil having 5 turns of radius 7cm. The coil lies in X-Y plane with its centre at the origin. Find the magnitude and direction of the magnetic dipole moment of the coil.<\/span><\/strong><\/p>\n

              Ans:\u00a0<\/span>\"\"<\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              Q7.<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Consider a tightly wound 100 turns coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil?<\/span><\/strong><\/p>\n

              Solution Since radius r = 10 cm = 0.1 m. The number of turns n = 100.\u00a0<\/span><\/p>\n

              So B=\u00a0<\/span>\"\"<\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              10 Magnetic field at a point on the axis of a circular coil carrying current:-<\/span><\/u><\/strong><\/p>\n

              Suppose a point P on the axis of a Circular coil carrying current I. Suppose two small current element\u00a0<\/span>\"\"\u00a0&\u00a0<\/span>\"\"\u00a0on coil. Then the Magnetic field due to these current Elements at P is dB & dB\u2019. As shown in fig.\u00a0<\/span><\/p>\n

              Clearly dB & dB\u2019 are equal. Resolving dB & dB\u2019 into rectangular components, we see that\u00a0<\/span>\"\"\u00a0&\u00a0<\/span>\"\"\u00a0\u00a0<\/span>are equal & opposite so they are cancelled out.\u00a0<\/span><\/p>\n

              So the total magnetic induction at P due to current through the whole circular coil is given<\/span>\u00a0\u00a0<\/span><\/p>\n

              B =\u0283dB Sin<\/span>\u2205<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u2013(i)<\/span><\/p>\n

              From Ampere circuital law, the magnetic induction due to\u00a0<\/span>\"\"\u00a0at P is<\/span>\u00a0\u00a0<\/span><\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              \"F:\\unitHere\u00a0<\/span>\"\"\u00a0is angle between r &\u00a0<\/span>\"\"\u00a0so 90<\/span>0<\/sup><\/span>.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              So<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>dB =<\/span>\"\"<\/p>\n

              Using in (i)\u00a0<\/span><\/p>\n

              B =<\/span>\u00a0\u00a0\u00a0<\/span>\u0283<\/span>\"\"\u00a0<\/span>\"\"<\/p>\n

              Here<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>sin<\/span>\u2205<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>&<\/span>\u00a0\u00a0\u00a0<\/span>\u0283 d<\/span>\"\"\u00a0=2<\/span>\"\"<\/p>\n

              \u00a0\u00a0\u00a0\u00a0<\/span>\u21d2 \u00a0\u00a0\u00a0<\/span>B =<\/span>\"\".<\/span>\"\"\u00a02<\/span>\"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

              If there are n turn in the coil\u00a0<\/span><\/p>\n

              then<\/span>\u00a0\u00a0\u00a0<\/span>B =\u00a0<\/span>\"\"<\/p>\n

              \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>at center\u00a0<\/span>\"\"\u00a0=0<\/span>\u00a0\u00a0\u00a0<\/span>\u21d2<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              B =\u00a0<\/span>\"\"<\/p>\n

              \"\"<\/p>\n

              \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If P point lies far away from the coil then\u00a0<\/span>\"\"\u00a0a so a can be neglected\u00a0<\/span><\/p>\n

              \u21d2<\/span> \u00a0\u00a0 B =<\/span>\"\"<\/p>\n

              =<\/span>\"\"<\/p>\n

              \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If P point lies at a distance equal to radius of the coil I,e. r = a we have<\/span><\/p>\n

              \u21d2\u00a0\u00a0 B=\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

              =<\/span>\"\"\u00a0<\/span><\/p>\n

              \"\"<\/p>\n

              11 The direction of magnet field due to circular coil:-<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong>\u00a0<\/strong><\/p>\n

              As we can see the magnetic lines form Close loop at the end of the circular coil & straight line at the center of the loop. The direction of these magnetic lines can be given by Right hand rule.<\/span><\/p>\n

              \"F:\\unit<\/p>\n

              \u00a0<\/span><\/p>\n

              \u00a0<\/span><\/p>\n

              \u00a0<\/span><\/p>\n

              \u00a0<\/span><\/strong><\/p>\n

              \u00a0<\/span><\/strong><\/p>\n

               <\/p>\n

              12 Right hand rule:-<\/span><\/u><\/strong>\u00a0<\/span><\/strong><\/p>\n

              \"\"Suppose the current is flowing through a circular conductor, if we imagine the fingers of the right hand curling in the direction of current, then the thumb will point in the direction of magnetic field.<\/span><\/p>\n

              \u00a0<\/span><\/strong><\/p>\n

              \u00a0<\/span><\/strong><\/p>\n

              13 Clock rule:-<\/span><\/u><\/strong>\u00a0<\/strong><\/p>\n

              According to clock rule if current moves in anti clock wise direction, the upper face of loop or coil is behave as north pole & when current moves in clock wise direction, then upper face behave as south pole.<\/span><\/p>\n

              \"F:\\unit<\/p>\n

              \u00a0<\/span><\/strong><\/p>\n

              \u00a0<\/span><\/strong><\/p>\n

              \u00a0<\/span><\/strong><\/p>\n

               <\/p>\n

              14 Ampere\u2019s circuital law:-<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong>\u00a0<\/span><\/u><\/strong><\/p>\n

              According to Ampere\u2019s circuital law, the line integral of magnetic field around a closed circuit is equal to\u00a0<\/span>\"\"\u00a0times the total current flowing through the circuit.<\/span><\/p>\n

              I,e.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span>=<\/span>\"\"I<\/span><\/p>\n

              Proof: –<\/span><\/u>\u00a0\u00a0\u00a0\u00a0<\/span><\/u><\/p>\n

              \"F:\\unitSuppose a Ampere\u2019s surface around a straight conductor carrying current I. Now the magnet field at P due to current I in conductor can be given by<\/span>\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              \"\"<\/p>\n

              Now the line integral of<\/span>\"\"\u00a0is<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

              \u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

              =\u00a0<\/span>\"\"\u00a0<\/span>\"\"<\/p>\n

              Here<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"= circumference of the circle<\/span><\/p>\n

              So\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

              =\u00a0<\/span>\"\"\u00a0<\/span>\"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

              This is required expression for Ampere circuital law.<\/span><\/p>\n

              \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Ampere\u2019s circular law holds for DC or steady current which does not change with time.<\/span><\/em><\/p>\n

              \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Ampere\u2019s circuital law for magnetic field is analogy to Gauss law in electrostatics.<\/span><\/em><\/p>\n

              \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Biot Savart\u2019s law & Ampere\u2019s circuital law are equivalent in same sense as Coulombs Law & Gauss theorem in electrostatics.<\/span><\/em><\/p>\n

              \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>For a loop that is not connected to the current<\/span><\/em>\u00a0\u00a0\u00a0<\/span><\/em>\"\"<\/p>\n

               <\/p>\n

              15 Applications of Ampere\u2019s Circuital law<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong>\u00a0<\/span><\/u><\/strong><\/p>\n

              \u00a0 <\/span>1. Magnetic field due to an infinite long current carrying wire:-<\/u><\/p>\n

              \"F:\\unitSuppose an infinite long straight conductor carrying current I. Now we have to calculate magnetic field at P on the Amperean loop of radius r. Consider small length\u00a0<\/span>\"\"\u00a0at loop, then magnetic field B at\u00a0<\/span>\"\"\u00a0can be given by Ampere\u2019s circuital law as\u00a0<\/span><\/p>\n

              \"\"\u00a0\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"<\/p>\n

              O r<\/span>\u00a0\u00a0<\/span>\"\"cos0<\/span>0<\/sup><\/span>\u00a0\u00a0\u00a0<\/span>=<\/span>\"\"<\/p>\n

              \u00a0<\/span>Or<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>B\u00a0<\/span>\"\"=<\/span>\"\"\u00a0<\/span><\/p>\n

              Here\u00a0<\/span>\"\"\u00a0= circumference of circle =\u00a0<\/span>\"\"\u00a0\u00a0<\/span><\/p>\n

              \uf0f0<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>B\u00a0<\/span>\"\"\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

              \uf0f0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>B<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>=<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>\"\"<\/p>\n

               <\/p>\n

              \"\"(b) Solenoid: –<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong><\/p>\n

              Solenoid means an insulated copper wire wounded closely in the form of a helix. The length of the solenoid is very large as compared to its diameter<\/span><\/em>.<\/span><\/p>\n

              Calculation of magnetic field inside a long straight solenoid:-<\/span><\/u><\/strong><\/p>\n

              \"F:\\unitSuppose a rectangular Amperean loop\u00a0<\/span>\"\"\u00a0of length<\/span>\"\". If N is the number of turns in the length<\/span>\"\", then total current through the loop equal to\u00a0<\/span>\"\".<\/span><\/p>\n

              Thus according to Amperes law<\/span>\u00a0\u00a0<\/span>\u00a0<\/span><\/strong><\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"\u00a0–(1)<\/span><\/p>\n

              \"RightOr<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u0283<\/span>abcda<\/sub><\/span>\u00a0<\/span>\"\"\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>\u0283<\/span>a<\/sub><\/span>b<\/sup><\/span>\u00a0<\/span>\"\"+<\/span>\u00a0\u00a0<\/span>\u0283<\/span>b<\/sub><\/span>c<\/sup><\/span>\u00a0<\/span>\"\"+<\/span>\u00a0\u00a0<\/span>\u0283<\/span>c<\/sub><\/span>d<\/sup><\/span>\u00a0<\/span>\"\"+ \u0283<\/span>d<\/sub><\/span>a<\/sup><\/span>\u00a0<\/span>\"\"<\/p>\n

              \u00a0<\/span>Here<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u0283<\/span>b<\/sub><\/span>c<\/sup><\/span>\u00a0<\/span>\"\"= \u0283<\/span>d<\/sub><\/span>a<\/sup><\/span>\u00a0<\/span>\"\"=<\/span>\u00a0\u00a0<\/span>\u0283\u00a0<\/span>\"\"cos<\/span>\"\"\u00a0\u00a0<\/span>= 0<\/span><\/p>\n

              Also<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u0283<\/span>c<\/sub><\/span>D<\/sup><\/span>\u00a0<\/span>\"\"\u00a0\u00a0<\/span>= 0 because CD lies outside the solenoid.<\/span><\/p>\n

              \"\"\u00a0= \u0283<\/span>a<\/sub><\/span>b<\/sup><\/span>\u00a0<\/span>\"\"\u00a0= B<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>..(2)<\/span><\/p>\n

              Comparing eq<\/span>n<\/sup><\/span>\u00a01 & 2 we get<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              \u00a0<\/span>B<\/span>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0<\/span>B =\u00a0<\/span>\"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>B =\u00a0<\/span><\/strong>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              Where n is the number of turns per unit length.<\/span><\/p>\n

              Q<\/span><\/strong>8.<\/span><\/strong>How is the magnetic field inside a given solenoid made strong?<\/span><\/strong><\/p>\n

                \n
              • By inserting a ferromagnetic substance inside the solenoid<\/span><\/li>\n
              • By increasing the amount of current through the solenoid<\/span><\/li>\n<\/ul>\n

                Q<\/span><\/strong>9. A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid?<\/span><\/strong><\/p>\n

                Solution<\/span>\u00a0\u00a0\u00a0<\/span>The number of turns per unit length is,\u00a0<\/span>\"\"\u00a0turns\/m\u00a0<\/span><\/p>\n

                The length\u00a0<\/span>\"\"\u00a0and radius<\/span>\"\"\u00a0<\/span><\/p>\n

                Thus,\u00a0<\/span>\"\"<\/p>\n

                \"\"Hence, we can use the long solenoid formula, namely,<\/span><\/p>\n

                \u00a0<\/span>B =\u00a0<\/span>\"\"\u00a0\u00a0<\/span><\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                 <\/p>\n

                c<\/span><\/u><\/strong>\u00a0\u00a0\u00a0<\/span><\/u><\/strong>Toroid or Toroidal Solenoid<\/span><\/u><\/strong>(not directly in cbse)<\/span><\/u><\/strong><\/p>\n

                A solenoid bent into the form of a closed ring is called a toroidal solenoid.<\/span><\/em><\/p>\n

                \"F:\\unitLet us consider a toroid having N number of turn equally spaced & let I is the amount of current flowing through them.<\/span><\/p>\n

                  \n
                • \u00a0\u00a0<\/span>Magnetic field at a point inside the core of the solenoid:-<\/u><\/li>\n<\/ul>\n

                  From Ampere circuital law<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                  \u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span><\/strong>=<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                  From N number of turns<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                  \u00a0<\/span>\"\"=\u00a0<\/span>\"\"<\/p>\n

                  Here\u00a0<\/span>\"\"=\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                  \u00a0<\/span>\uf0f0<\/span>\u00a0\u00a0\u00a0<\/span>B<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                  \"\"<\/p>\n

                  Or<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>B =\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/p>\n

                  Where n =\u00a0<\/span>\"\"\u00a0= number of turns per unit length, Thus magnetic is same as that of solenoid.<\/span><\/p>\n

                    \n
                  • \u00a0\u00a0<\/span>\u00a0Magnetic field inside the toroid:-<\/u><\/li>\n<\/ul>\n

                    \u00a0<\/span>As inside a hollow conductor I = 0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                    \u00a0<\/span>\uf0f0<\/span>\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                    =\u00a0<\/span>\"\"<\/p>\n

                    \"\"\uf0f0<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>B=0<\/span><\/p>\n

                    Thus Magnetic field inside the toroid is zero<\/span><\/p>\n

                      \n
                    • \u00a0\u00a0<\/span>\u00a0 \u00a0Magnetic field outside the solenoid.<\/u><\/li>\n<\/ul>\n

                      For any point outside the tortoid, the current threading the loop L\u2019 through current at the point is zero so B = 0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      \uf0f0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0= 0<\/span><\/p>\n

                      This is the condition for ideal toroid in which the turns of the wire are very closely spaced & the magnetic field is within the toroid. For outside B = 0. I,e. magnetic field does not comes outside as in solenoid.<\/span><\/u><\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/em><\/strong><\/p>\n

                      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em><\/strong><\/p>\n<\/div>\n

                      \n

                      16. Motion of a charge particle in a uniform electric field:-<\/span><\/u><\/strong><\/p>\n

                      Let us consider a charge particle + q moving with\u00a0<\/span>\"\"\u00a0velocity in the electric field then charge particle will experience a electric force.<\/span><\/p>\n

                      \"F:\\unitAs\u00a0<\/span>\"\"\u00a0= q<\/span>\"\"-(1)<\/span><\/p>\n

                      Due to this electric force charge particle starts accelerating with a force.<\/span><\/p>\n

                      \"\"\u00a0= m<\/span>\"\"-(2)<\/span><\/p>\n

                      Comparing eq<\/span>n\u00a0<\/sup><\/span>(1) & (2) we get<\/span><\/p>\n

                      m<\/span>\"\"\u00a0= q<\/span>\"\"<\/p>\n

                      \"\"\u00a0=\u00a0<\/span>\"\"-(3)<\/span><\/p>\n

                      Now let the time taken by the charge particle to move\u00a0<\/span>\"\"\u00a0distance is\u00a0<\/span><\/p>\n

                      t =\u00a0<\/span>\"\"<\/p>\n

                      Similarly the distance travelled along y axis is<\/span><\/p>\n

                      \"\"\u00a0=\u00a0<\/span>\"\"t +\u00a0<\/span>\"\"\"\"t<\/span>2<\/sup><\/span><\/p>\n

                      Here<\/span>\"\"\u00a0= 0<\/span>also<\/span>\"\"\u00a0=\u00a0<\/span>\"\"&<\/span>\"\"<\/p>\n

                      So<\/span>y<\/span>=<\/span>\"\"2<\/sup><\/span><\/p>\n

                      =<\/span>\"\"<\/p>\n

                      Or<\/span>\"\"<\/p>\n

                      where<\/span>k =<\/span>\"\"<\/p>\n

                      Or<\/span>y\u00a0<\/span>\"\"\u00a0x<\/span>2<\/sup><\/span><\/p>\n

                      Which is an eq<\/span><\/em>n<\/sup><\/span><\/em>\u00a0of parabola, hence we can say that particle will follow\u00a0<\/span><\/em>parabolic path.<\/span><\/em><\/strong><\/p>\n

                       <\/p>\n

                      17. Motion of charge particle in uniform magnetic field:-<\/span><\/u><\/strong><\/p>\n

                      Let us suppose a charge particle +q enter in a magnetic field B at origin O with a velocity<\/span>\"\", making an angle \u03b8 with magnetic field.\u00a0<\/span><\/p>\n

                      \"F:\\unitNow resolving\u00a0<\/span>\"\"\u00a0into two rectangular components we have\u00a0<\/span>\"\"1<\/sub><\/span>\u00a0=\u00a0<\/span>\"\"cos\u03b8 along B and\u00a0<\/span>\"\"2\u00a0<\/sub><\/span>=\u00a0<\/span>\"\"\u00a0sin\u03b8 along perpendicular to magnetic field,\u00a0<\/span><\/p>\n

                      The force on the charge particle due to\u00a0<\/span>\"\"2<\/sub><\/span>\u00a0is\u00a0<\/span><\/p>\n

                      \"\"<\/p>\n

                      Or<\/span>\"\"<\/p>\n

                      Or<\/span>\"\"<\/p>\n

                      The direction of this force can be obtained by R.H.R & is perpendicular to plane containing<\/span>\"\".\u00a0<\/span><\/p>\n

                      Here motion of the charge particle is perpendicular to applied force so work done by this force is zero.<\/span><\/p>\n

                      Hence this force provides no motion to the charge particle but gives the circular motion. i,e. centripetal force. So we may write<\/span><\/p>\n

                      \"\"\u00a0=<\/span>\"\"<\/p>\n

                      or<\/span>\"\"=<\/span>\"\"<\/p>\n

                      The angular velocity of the particle can be given as<\/span><\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      The frequency of rotation of the particle in magnetic field will Becomes<\/span><\/p>\n

                      \"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                      =<\/span>\"\"<\/p>\n

                      So the time period<\/span><\/p>\n

                      T =<\/span>\"\"\u00a0=<\/span>\"\"<\/p>\n

                      For component\u00a0<\/span>\"\"1<\/sub><\/span>\u00a0=\u00a0<\/span>\"\"cos\u03b8 the force is<\/span><\/p>\n

                      \"\"<\/p>\n

                      =\u00a0<\/span>\"\"\u00a0= 0<\/span><\/p>\n

                      Here there is no any force due to\u00a0<\/span>\"\"\u00a0so the charge particle moves linearly due to component<\/span>\"\".<\/span><\/p>\n

                      Thus by the combined effect of\u00a0<\/span><\/em>\"\"&<\/span><\/em>\"\"\u00a0charge particle move linear as well as circular so its path becomes\u00a0<\/span><\/em>helical<\/span><\/em><\/strong>. The total distance travelled by the particle is\u00a0<\/span><\/p>\n

                      \"\"<\/p>\n

                      Q10. An electron and a proton are moving along the same direction with the same kinetic energy. They enter a uniform magnetic field acting perpendicular to their velocities. The dependence of radius of their paths on their masses is:<\/span><\/strong><\/p>\n

                      A.<\/span>\"\"B.<\/span><\/strong>\"\"C.<\/span>\"\"D.<\/span>\"\"<\/p>\n

                      Q11.An alpha particle is projected with velocity<\/span><\/strong>\"\"into a region in which magnetic field\u00a0<\/span><\/strong>\"\"exists. Calculate the acceleration of the particle in the region, charge to mass ratio for alpha particle is\u00a0<\/span><\/strong>\"\"<\/p>\n

                      Ans:<\/span>\"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      Q12.<\/span><\/strong>What is the radius of the path of an electron (mass 9 \u00d7 10<\/span><\/strong>-31<\/sup><\/span><\/strong>\u00a0kg and charge 1.6 \u00d7 10<\/span><\/strong>\u201319<\/sup><\/span><\/strong>\u00a0C) moving at a speed of 3 \u00d710<\/span><\/strong>7<\/sup><\/span><\/strong>\u00a0m\/s in a magnetic field of 6 \u00d7 10<\/span><\/strong>\u20134<\/sup><\/span><\/strong>\u00a0T perpendicular to it?<\/span><\/strong>What is its frequency?<\/span><\/strong>Calculate its energy in<\/span><\/strong>keV.<\/span><\/strong>( 1 eV = 1.6 \u00d7 10<\/span><\/strong>\u201319<\/sup><\/span><\/strong>\u00a0J).<\/span><\/strong><\/p>\n

                      Solution<\/span>As<\/span>\"\"<\/p>\n

                      =\u00a0<\/span>\"\"<\/p>\n

                      = 26 \u00d7\u00a0<\/span>\"\"\u00a0m = 26 cm<\/span><\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"Hz<\/span>=2 MHz.<\/span><\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                       <\/p>\n

                      18. Lorentz force<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong><\/p>\n

                      The total force experienced by a charge particle in electric & magnetic field both is called Lorentz force.\u00a0<\/span><\/p>\n

                      The force experienced by charge particle in electric field is<\/span><\/p>\n

                      \"\"=<\/span>q<\/span>\"\"–(1)<\/span><\/p>\n

                      & the force experienced in magnetic field is<\/span><\/p>\n

                      \"\"\u2013(2)<\/span><\/p>\n

                      Thus the total force (Lorenz force) experienced by charge particle is<\/span><\/p>\n

                      \"\"=<\/span>\"\"+\u00a0<\/span>\"\"<\/p>\n

                      Or<\/span>\"\"= q<\/span>\"\"\u00a0+\u00a0<\/span>\"\"<\/p>\n

                      Or<\/span>F =<\/span><\/strong>q<\/span><\/strong>\"\"+<\/span><\/strong>\"\"<\/p>\n

                      Special cases:-<\/span><\/u><\/p>\n

                        \n
                      1. When all \"\",\"\"\u00a0 are collinear.\u00a0 In this case \"\"\u00a0= 0 because \"\"<\/li>\n<\/ol>\n

                        So the total force on the charge particle will be\u00a0<\/span><\/p>\n

                        \"\"\u00a0= q<\/span>\"\"<\/p>\n

                          \n
                        1. When\"\",\"\"\u00a0are mutually perpendiculars to each other, In this case \"\"\u00a0\"\"\u00a0 are such that<\/li>\n<\/ol>\n

                          \"\"\u00a0=\u00a0<\/span>\"\"\u00a0+\u00a0<\/span>\"\"= 0,\u00a0<\/span><\/p>\n

                          then\u00a0<\/span>\"\"\u00a0=<\/span>\"\"=0<\/span><\/p>\n

                          It means the particle will pass through the field without any change in velocity. The concept is use in velocity Selector to obtain a charged beam having a definite velocity.<\/span><\/p>\n

                           <\/p>\n

                          19.\u00a0<\/span><\/u><\/strong>Motion of charge particle in perpendicular magnetic & Electric field:-\u00a0<\/span><\/u><\/strong>Or<\/span><\/strong>\u00a0Velocity Selector:-<\/span><\/u><\/strong><\/p>\n

                          In a Velocity Selector there are both electric field and magnetic field are perpendicular to each other. They are also perpendicular to direction of motions of electrons.\u00a0<\/span><\/p>\n

                          \"F:\\unit
                          Velocity Selector:<\/figcaption><\/figure>\n

                          The electrons which have equal and opposite electric and magnetic field may pass through the velocity selector.\u00a0<\/span><\/p>\n

                          The velocity of electrons in this case may be given as<\/span><\/p>\n

                          \"\"<\/p>\n

                          \"\"<\/p>\n

                           <\/p>\n

                          20. Force on a current carrying conductor placed in magnetic field.<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong><\/p>\n

                          \"F:\\unitConsider a conductor PQ of length\u00a0<\/span>\"\"\u00a0and area<\/span>\"\". Suppose current flows in the conductor along + ve y axis & magnetic field is applied along +ve z axis. So the velocity\u00a0<\/span>\"\"\u00a0of electrons will be along-ve y axis.\u00a0<\/span><\/p>\n

                          The magnetic force experienced by the electron may be given as<\/span><\/p>\n

                          \"\"<\/p>\n

                          If there are n numbers of electrons per unit volume then total numbers of electrons are<\/span>\"\"<\/p>\n

                          So the total force experienced by the electrons is<\/span><\/p>\n

                          \"\"<\/p>\n

                          \"\"(1)<\/span><\/p>\n

                          As we know<\/span><\/p>\n

                          \"\"<\/p>\n

                          Or<\/span>\"\"<\/p>\n

                          \"\"<\/p>\n

                          (As\u00a0<\/span>\"\"are in opposite direction so<\/span>-ve sign is taken,)\u00a0<\/span><\/p>\n

                          Now from e<\/span>\"\"\u00a0(1)<\/span>\"\"<\/p>\n

                          \"\"<\/p>\n

                          Or<\/span>\"\"<\/p>\n

                          where\u00a0<\/span>\"\"\u00a0is the direction between<\/span>\"\".<\/span><\/p>\n

                          Direction of force:-\u00a0<\/span><\/u><\/p>\n

                          The direction of force is given by Fleming\u2019s left hand rule or Right hand thumb rule.<\/span><\/p>\n

                          Special cases:-<\/span><\/u><\/p>\n

                            \n
                          1. If\u00a0\u00a0\u00a0\u00a0 \u03b8\u00a0 =\u00a0 0\u00b0\u00a0\u00a0 or\u00a0\u00a0 180\"\"<\/li>\n
                          2. then\u00a0\u00a0\u00a0 sin0 = 0\"\"\u00a0F = 0<\/li>\n<\/ol>\n

                            Hence conductor placed parallel to direction of magnetic field does not experience any force.<\/span><\/p>\n

                              \n
                            1. If \"\"=\u00a0 90\"\"sin 90 = 1\"\"<\/li>\n<\/ol>\n

                              \"\"<\/p>\n

                              When conductor is placed perpendicular to magnetic field it will experience maximum force.<\/span><\/p>\n

                              Q13. The horizontal component of the earth\u2019s magnetic field at a certain place is 3.0 \u00d710<\/span><\/strong>\u20135<\/sup><\/span><\/strong>T and the direction of the field is from geographic south to the north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west; (b) south to north?<\/span><\/strong><\/p>\n

                              Solution\u00a0<\/span>\"\"<\/p>\n

                              The force per unit length is\u00a0<\/span>\"\"<\/p>\n

                              (a)When the current is flowing from east to west, \u03b8 = 90\u00b0\u00a0<\/span><\/p>\n

                              Hence,\u00a0<\/span>\"\"<\/p>\n

                              This is larger than the value 2\u00d710\u20137 Nm\u20131 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth\u2019s magnetic field and other stray fields while standardizing the ampere. The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.<\/span><\/p>\n

                              (b) When the current is flowing from south to north,\u00a0<\/span>\"\"\u00a0so<\/span>\"\"\u00a0hence there is no force on the conductor<\/span><\/p>\n

                               <\/p>\n

                              21. Expression for the force between two Parallel current carrying wire<\/span><\/u><\/strong>m.imp.<\/sup><\/span><\/u><\/strong><\/p>\n

                              \"F:\\unit\"F:\\unit<\/p>\n

                               <\/p>\n

                               <\/p>\n

                              Consider two long parallel wires AB & CD carrying current<\/span>\"\". Let r be the separation between them. The magnetic field produced by current\u00a0<\/span>\"\"\u00a0at any point on wire CD is<\/span><\/p>\n

                              \"\"<\/p>\n

                              This field acts perpendicular to the wire CD and points into the plane of the paper .It exerts a force given as<\/span><\/p>\n

                              \"\"<\/p>\n

                              According to Fleming\u2019s left hand rule, this force acts at right angle to CD, toward AB in the plane of the paper.<\/span><\/p>\n

                              Similarly an equal force is exerted on the wire AB by the field of wire CD.<\/span><\/p>\n

                              As\u00a0<\/span>\"\"<\/p>\n

                              Again according to Fleming\u2019s left hand rule, this force acts at right angle to AB, toward CD in the plane of the paper.\u00a0<\/span>Thus when the current in the two wires are in same direction, the force is attractive & when current are in opposite direction, force is repulsive.<\/span><\/em><\/p>\n

                              Imp.\u00a0<\/sup><\/span><\/em><\/strong>The ampere is that constant current which\u2014if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 meter apart in vacuum\u2014would produce between these conductors a force equal to<\/span><\/em>\"\"Newton\u2019s per meter of length.<\/span><\/em><\/p>\n

                              Q14.<\/span><\/strong>\"\"Two long parallel straight conductors are placed 12cm apart in air. They carry equal currents of 3A each. Find the magnitude and direction of the magnetic field at a point midway between them (drawing a figure) when the currents in them flow in opposite directions.\u00a0<\/span><\/strong><\/p>\n

                              Solution\u00a0<\/span>\"\"<\/p>\n

                              \"\"<\/p>\n

                              \"\"(<\/span>\"\")<\/span><\/p>\n

                              \"\"<\/p>\n

                              \"\"\u00a0T<\/span>\"\"<\/p>\n

                               <\/p>\n

                              22. TExpression for the force between two Parallel current carrying wire:-<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong><\/p>\n

                              .Let us consider a current carrying rectangular coil of sides\u00a0<\/span>\"\"\u00a0such that area<\/span>\"\".\u00a0<\/span><\/p>\n

                              Suppose this coil is placed at an angle \u03b8 with magnetic field B.\u00a0<\/span><\/p>\n

                              Now as we know that when a current carrying conductor placed in magnetic field experience a force,\u00a0<\/span><\/p>\n

                              So force on side PQ<\/span><\/p>\n

                              \"F:\\unit\"\"(<\/span>\"\")\u00a0<\/span><\/p>\n

                              \"\"= IlBsin90<\/span>0<\/sup><\/span>\u00a0= I<\/span>\"\"\u2026..(1)<\/span><\/p>\n

                              Again Force on side QR<\/span><\/p>\n

                              =\u00a0<\/span>\"\"(<\/span>\"\")\u00a0<\/span>\"\"<\/p>\n

                              \"\"\u00a0=\u00a0<\/span>\"\"\u2026\u2026\u2026..\u2026..(2)<\/span><\/p>\n

                              And force on side RS<\/span><\/p>\n

                              \"\"(<\/span>\"\")<\/span>\"\"<\/p>\n

                              \"\"=<\/span>\"\"<\/p>\n

                              = I<\/span>\"\"\u2026….(3)<\/span><\/p>\n

                              Similarly Force on side SP<\/span><\/p>\n

                              \"\"(<\/span>\"\")<\/span><\/p>\n

                              \"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                              \"\"\u00a0=\u00a0<\/span>\"\"\u00a0\u2026\u2026.(4)<\/span><\/p>\n

                              According to Fleming\u2019s left hand rule the Forces\u00a0<\/span>\"\"\u00a0&\u00a0<\/span>\"\"\u00a0are equal and act along the axis of the coil in opposite direction so exert no any torque they are cancelled out.\u00a0<\/span><\/p>\n

                              While the force\u00a0<\/span>\"\"\u00a0&\u00a0<\/span>\"\"\u00a0will exert a torque on the coil given as<\/span><\/p>\n

                              \"\"<\/p>\n

                              \"\"<\/p>\n

                              \"\"<\/p>\n

                              \"\"(<\/span>\"\")<\/span><\/p>\n

                              0r<\/span>\"\"<\/p>\n

                              Where<\/span>\"\"\u00a0= magnetic dipole moment\u00a0<\/span><\/p>\n

                              Q15. A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T, normal to the plane of the coil. If the current in the coil is 3.0 A, calculate the<\/span><\/strong><\/p>\n

                              1. Total torque on the coil,\u00a0<\/span><\/strong><\/p>\n

                              2. Total force on the coil.\u00a0<\/span><\/strong><\/p>\n

                              3. Average force on each electron in the coil, due to the magnetic field. Assume the area of cross – section of the wire to be\u00a0<\/span><\/strong>\"\"\u00a0and the free electron density is\u00a0<\/span><\/strong>\"\".<\/span><\/strong><\/p>\n

                              Ans:\u00a0<\/span><\/strong><\/p>\n

                                \n
                              1. \"\"<\/li>\n
                              2. \"\"<\/li>\n
                              3. \"\"\"\"<\/li>\n<\/ol>\n

                                 <\/p>\n

                                23<\/span><\/strong>Moving coil Galvanometer:-<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong><\/p>\n

                                A<\/span><\/em>\u00a0Galvanometer is a device which is used to detect the current in a circuit<\/span>.<\/span><\/em><\/p>\n

                                Principle:-<\/span><\/u><\/p>\n

                                The Galvanometer is based on the principal that a current carrying coil placed in external magnetic field experience a torque<\/span><\/em>.<\/span><\/p>\n

                                Construction:-<\/span><\/u> –<\/p>\n

                                A moving coil Galvanometer consist of<\/span><\/p>\n<\/div>\n

                                \"\"<\/p>\n

                                 <\/p>\n

                                 <\/p>\n

                                 <\/p>\n

                                 <\/p>\n

                                 <\/p>\n

                                 <\/p>\n

                                 <\/p>\n

                                \n
                                  \n
                                1. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 A coil PQRS consist of a large number of turns of fine insulated copper wire wounded over a non magnetic metallic frame. The coil is suspended from a movable torsion head H by mean of a fine phosphor bronze strip. The lower end of coil is connected to fine spiral spring S\u2019. An arrangement of strong north & South Pole is fixed on both sides of the coil.<\/li>\n
                                2. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 A soft iron core L of spherical or cylindrical shape is placed between the coils.<\/li>\n
                                3. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0A circular Mirror M is attached to the Phosphor bronze strip, to measure the deflection of the coil.<\/li>\n
                                4. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0The torsion head is connected to a binding terminal T1<\/sub><\/span> & spring S\u2019 is connected to another binding terminal T2<\/sub><\/span>.<\/li>\n<\/ol>\n

                                  Theory:-<\/span><\/u><\/p>\n

                                  Let N is number turns in coil, A is area of coil, B is magnetic field I is the amount of current following through the coil, then torque acting on the coil is\u00a0<\/span><\/p>\n

                                  \"\"<\/p>\n

                                  Here \u03b8 is angle between normal to the plane of coil & applied magnetic field.<\/span><\/p>\n

                                  Since the field is\u00a0<\/span>radial<\/span><\/strong><\/p>\n

                                  i,e<\/span>\u03b8 = 90<\/span>0\u00a0<\/sup><\/span>\"\"<\/p>\n

                                  \"\"<\/p>\n

                                  Due to this torque the phosphor bronze wire suffer twist, if \u03b8 is angle of twist & k is moment of restoring couple per unit angular twist then moment of restoring couple = k \u03b8<\/span><\/p>\n

                                  In equilibrium<\/span><\/p>\n

                                  \"\"<\/p>\n

                                  Or<\/span>\"\"<\/p>\n

                                  Here\u00a0<\/span>\"\"\u00a0= K = a another constant\u00a0<\/span><\/p>\n

                                  So<\/span>I\u00a0<\/span><\/strong>\"\"\u00a0\u03b8<\/span><\/strong><\/p>\n

                                  Hence current through the coil is directly proportional to the deflection of the coil.<\/span><\/em><\/p>\n

                                   <\/p>\n

                                  Sensitivity of a moving coil Galvanometer<\/span><\/u><\/strong><\/p>\n

                                  A galvanometer is said to be more sensitive if it show large deflection even for a small change in current is passed through it.<\/span><\/p>\n

                                    \n
                                  1. Current sensitivity: – it is defined as the deflection produced in the galvanometer when a unit current is passed through it.<\/li>\n<\/ol>\n

                                    I,e.\u00a0<\/span>Current sensitivity<\/span><\/p>\n

                                    \"\"<\/p>\n

                                      \n
                                    1. Voltage sensitivity: – It is defined as the deflection produced in the galvanometer when a unit potential difference is applied across it.<\/li>\n<\/ol>\n

                                      Voltage sensitivity <\/span>V<\/span>s<\/sub><\/span>=<\/span>\"\"=<\/span>\"\"<\/p>\n

                                      =<\/span>\"\"<\/p>\n

                                      Clearly voltage sensitivity =<\/span>\"\"<\/p>\n

                                      The sensitivity of a moving coil galvanometer can be increased by\u00a0<\/span><\/em><\/p>\n

                                      \uf0fc<\/span>By increasing number of turns in the coil.<\/span><\/p>\n

                                      \uf0fc<\/span>By increasing the magnetic field.<\/span><\/p>\n

                                      \uf0fc<\/span>By increasing the area of the oil.<\/span><\/p>\n

                                      \uf0fc<\/span>By decreasing the value of torsion constant R<\/span><\/p>\n

                                      Advantage of moving coil Galvanometer:-<\/span><\/u><\/p>\n

                                        \n
                                      • It tells about the deflection due to current in a circuit.<\/span><\/li>\n<\/ul>\n

                                        Disadvantage of a moving coil Galvanometer:<\/span><\/u><\/p>\n

                                          \n
                                        • Its sensitivity cannot be changed at will.<\/span><\/li>\n
                                        • It is damaged by overloading.\u00a0<\/span><\/li>\n<\/ul>\n

                                           <\/p>\n

                                          Q16.What is the importance of a radial magnetic field and how is it produced?<\/span><\/strong><\/p>\n

                                          In a radial magnetic field magnetic torque remains maximum for all positions of the coils. It is produced due to cylindrical pole pieces and soft iron core.<\/span><\/p>\n

                                          24 Conversion of a Galvanometer into an ammeter<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong>:-<\/span><\/u><\/strong><\/p>\n

                                          (An ammeter is a device which is used to measure the current in a circuit, as current flow in wire so it is connected in series.\u00a0<\/span><\/em>An ammeter is formed by connecting a low resistance (shunt) in parallel to Galvanometer. The value of shunt depends upon the current which is to be measured.)<\/span><\/p>\n

                                          \"F:\\unitAs shunt & Galvanometer are connected in parallel\u00a0<\/span><\/p>\n

                                          So<\/span>\"\"G = (I \u2013\u00a0<\/span>\"\") S<\/span>\"\"<\/p>\n

                                          Or<\/span>\"\"<\/p>\n

                                          So by connecting a shunt across galvanometer, we get an ammeter of desired range,\u00a0<\/span><\/p>\n

                                          Moreover<\/span>\"\"<\/p>\n

                                          \"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                                          As the deflection in galvanometer is directly proportional to\u00a0<\/span>\"\"\u00a0& hence perpendicular so the current can be measured. Also the effective resistance is<\/span><\/p>\n

                                          R –\u00a0<\/span>\"\"\u00a0< S<\/span><\/p>\n

                                            \n
                                          • An ideal ammeter has zero resistance.<\/span><\/u><\/em>(imp:-MCQ)<\/span><\/strong><\/li>\n
                                          • Higher the range of ammeter to be prepared lower should be the value of shunt.<\/span><\/li>\n
                                          • The range of ammeter can be increased but cannot be decreased.<\/span><\/li>\n<\/ul>\n

                                            Q17. A galvanometer has a resistance of 30\u2126. It gives full scale deflection with a current of 2 mA. Calculate the value of the resistance needed to convert it into an ammeter of range 0-0.3 A.<\/span><\/strong><\/p>\n

                                            Ans: As\u00a0<\/span>\"\"<\/p>\n

                                            Substitution so\u00a0<\/span>\"\"<\/p>\n

                                            25 Conversion of a galvanometer into a Voltmeter:-<\/span><\/u><\/strong>\u00a0m.imp<\/sup><\/span><\/u><\/strong><\/p>\n

                                            (Voltmeter is a device which is used to measure the potential difference between two points in a circuit so connected in parallel.\u00a0<\/span><\/em>A voltmeter is formed by connecting a high resistance in series a Galvanometer.)<\/span><\/p>\n

                                            \"F:\\unitThe total resistance in the Circuit is\u00a0<\/span>\"\"<\/p>\n

                                            Now by ohm\u2019s Law<\/span><\/p>\n

                                            \"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                                            Or<\/span>\"\"<\/p>\n

                                            \"\"<\/p>\n

                                            So by connecting a high resistance in Series with G, we can get a voltmeter of desire range. As deflection is proportional to\u00a0<\/span>\"\"\u00a0& hence V. So the scale can be arranged to measure potential difference. Hence a voltmeter is a high resistance galvanometer.<\/span><\/p>\n

                                            Its effective resistance is\u00a0<\/span>\"\"<\/p>\n

                                              \n
                                            • An ideal voltmeter should have infinite resistance.<\/span>(imp:-MCQ)<\/span><\/em><\/strong><\/li>\n<\/ul>\n

                                              Why is it that while using a moving coil galvanometer as a voltmeter a high resistance in series is required whereas in an ammeter a shunt is used?<\/span><\/strong><\/p>\n

                                              Reason:<\/span><\/p>\n

                                              Voltmeter: This<\/span>ensures<\/span>that<\/span>a<\/span>very<\/span>low<\/span>current<\/span>passes<\/span>through<\/span>the<\/span>voltmeter<\/span>and<\/span>hence<\/span>does<\/span>not<\/span>change<\/span>(much)<\/span>the <\/span>original <\/span>potential difference to be measured.<\/span><\/p>\n

                                              Ammeter: This ensures that the total resistance of the circuit does not change much and the current flowing remains (almost) at its original value.<\/span><\/p>\n

                                              Example8. In the circuit the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance\u00a0<\/span><\/strong>\"\"; (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance\u00a0 \u00a0<\/span><\/strong>r s = 0.02<\/span><\/strong>\u2126; (c) is an ideal ammeter with zero resistance?<\/span><\/strong><\/p>\n

                                              Solution (a) Total resistance in the circuit is<\/span>\"\"\u00a0.<\/span><\/p>\n

                                              Hence<\/span>\"\"<\/p>\n

                                                \n
                                              1. \u00a0 \u00a0 \u00a0 \u00a0Resistance of the galvanometer converted to an ammeter is,<\/li>\n<\/ol>\n

                                                \"\"<\/p>\n

                                                Total resistance in the circuit is,\u00a0<\/span>\"\"<\/p>\n

                                                Hence,<\/span>\"\"<\/p>\n

                                                  \n
                                                1. For the ideal ammeter with zero resistance,<\/li>\n<\/ol>\n

                                                  \"\"<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

                                                  Chapter\u20134:\u00a0\u00a0Moving Charges and Magnetism 1. Electromagnetism The branch of physics which deals with the study of\u00a0magnetism due to electricity\u00a0is called electromagnetism. The relationship between electricity & Magnetism was firstly given by Oersted which is as below   2. Oersted experiments:-\u00a0 As shown in fig, Oersted found a relation between current & magnetism, when a magnetic […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"default","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"class_list":["post-2163","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_hostinger_reach_plugin_has_subscription_block":false,"_hostinger_reach_plugin_is_elementor":false,"uagb_featured_image_src":{"full":false,"thumbnail":false,"medium":false,"medium_large":false,"large":false,"1536x1536":false,"2048x2048":false},"uagb_author_info":{"display_name":"sandeep.soni484@gmail.com","author_link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/author\/sandeep-soni484gmail-com\/"},"uagb_comment_info":0,"uagb_excerpt":"Chapter\u20134:\u00a0\u00a0Moving Charges and Magnetism 1. Electromagnetism The branch of physics which deals with the study of\u00a0magnetism due to electricity\u00a0is called electromagnetism. The relationship between electricity & Magnetism was firstly given by Oersted which is as below   2. Oersted experiments:-\u00a0 As shown in fig, Oersted found a relation between current & magnetism, when a magnetic…","_links":{"self":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages\/2163","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/comments?post=2163"}],"version-history":[{"count":11,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages\/2163\/revisions"}],"predecessor-version":[{"id":2904,"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/pages\/2163\/revisions\/2904"}],"wp:attachment":[{"href":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/wp-json\/wp\/v2\/media?parent=2163"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}