Chapter–4:  Moving Charges and Magnetism

1. Electromagnetism

The branch of physics which deals with the study of magnetism due to electricity is called electromagnetism. The relationship between electricity & Magnetism was firstly given by Oersted which is as below

 

2. Oersted experiments:- 

As 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, 

When the direction of flow of current is reversed then the deflection is also reverses, the direction of deflection can be given by Ampere’s swimming rule.

 

Ampere’s swimming rule:-

According 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.

 

3. Magnetic field:-

The space around a magnet or a current carrying conductor in which its magnetic effect can be experienced is called magnetic field. It is a vector quantity & denoted by.

Force on a moving charge in a magnetic field or Magnetic Lorentz force.^m.imp 

Suppose a +q charge moving in a magnetic field  with velocity. Then  is the magnetic force experienced by the charge depends on

 F_m     q                        (i)

F_m   sinθ                   (ii)

F_m  B                        (iii)

Combining eqⁿs we get           

F_m        q B Sinθ 

Or    F_m      =     K q  B Sinθ 

Here       K = 1   

So Fm       ₌       q B Sinθ

Or in vector form                  

          ₌      q ( ) 

The direction of  can be given by Right hand rule.  

I,e.   is perpendicular to plane containing both  & . 

Also direction of    is given by Fleming left hand rule.

 

Definition of B:-  

If  =1,     q = 1    &      

  Then     

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.

 

Special cases:-

(i)-     If θ = 0⁰ or 180⁰   then   sin0⁰ =  0    ⇒  0

Hence a charge particle moving along or opposite to direction of magnetic field experience no any force.

(ii)-     If      

Means a rest charge in magnetic field experienced no any force.

(iii)-    If  θ = 90⁰  ⇒   sin 90⁰  =   max        maximum.

Hence a charge particle moving perpendicular to external magnetic field experience maximum force.

 

Unit of magnetic field B:    

SI unit of B is Tesla   or   Weber/metre² 

As        B =          

Or   1 Tesla =  

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^-1 experience a force of one Newton.

 

Dimensional formula: –  

B =     

=    

= [MA^-1 T^-2]

In c,g,s system the unit of magnetic field is gauss G.                  

1 T =   10⁴G

 

4. Right hand rule or right hand screw rules:- 

According to right hand rule if we curl the finger of our right hand from  to  then thumb will give the direction of _m.

 

5 Fleming’s left hand rule:- ^m.imp 

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.

 

6 Biot Savart’s Law:- ^m.imp 

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’s Law.

According to this law, the magnitude of magnetic field d at 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

  

Combining all eqⁿs we get    

 

Or        =   

Where  K     =  = 10^-7  TmA^-1      

 Here u₀ called permeability of free space. 

So Biot Savart’s law become 

    =  

In vector form                  

     =   

The direction of dB can be given by right hand rule. 

^Imp. In present case the direction of  is perpendicular to plane of paper containing both  &  in inward direction; if P point lies in L.H.S then in outwards direction.

Special cases:-

1.                   If   

2.               If 

  I,e. magnetic field at a point perpendicular to current element is maximum

     This law is applicable for infinite small conductor.

     This law is similar to coulomb’s law in electrostatics.

     This law is difficult to verify experimentally because it is difficult to construct infinite small conductor.

 

7 Biot Savart’s law V/s Coulomb’s Law:-

As according to Coulomb’s Law  

  =   

& from Biot Savart’s law                

dB =   

On comparing above eqⁿs we can give similarity & dissimilarity between both.

Similarity

1. Both the law obey inverse square law (α )
2. Both the law obeys principle of Superposition.

Dissimilarity:-

1. The source of magnetic field is current while source of electric field is charged.
2. Electric field exists due to both rest & motion of charge while magnetic field exists due to motion of charge.
3. The magnetic field is angle dependent while electric field is not.

8 Magnetic field due to a long straight Conductor carrying current:- ^m.imp

Suppose a point P having a distance a from the straight conductor carrying current I, then according to Biot savart Law, the magnetic field   at P due to small current element ,   r distance apart is

         –(1)

Now from Δ COP      

               

Or               (i)

Also  

    (ii)

As    tan  = 

⇒    = a tan

Differentiating both side d =a sec² .d          (iii)

Using all values in eqⁿs (1) we get   

 =

Or         = 

To calculate complete magnetic field integrating both sides we get       

B =  

B =  

B =  [Sin₂ – Sin(-∅₁)]

Or   B =     [Sin∅₁ + Sin∅₂]   

The direction of B can be given by Right hand rule.

Special cases:-

1.                 If the conductor is infinite long & point P lies near to the dipole then

B =  [sin90⁰ + sin90⁰]          

Or B = =   

2.                If the conductor is infinite long but P lies near end y then  ∅₁=  90,  ∅₂ =0

B =  [sin90 + sin0]    

 =  

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: 

Ans:   

 So 

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?

Solution we find that there is an upward force F, of magnitude, for mid-air suspension, this must be balanced by the force due to gravity:      

⇒       

0.65 T

Note that it would have been sufficient to specify, the mass per unit length of the wire. The earth’s magnetic field is approximately 4 × 10^–5T and we have ignored it.

 Q3.  An element    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. ∆x = 1 cm. 

Solution  here  

 ,

 

⇒  

  

The direction of the field is in the +z-direction. 

As    

 

9 Magnetic field At the Center Of The Circular Coil Carrying current:- ^m.imp

According to Biot Savart Law, the magnetic field at the center of circular coil carrying current can be given as   

         –(1)      

 (Here  is angle between current element Id & r. )

As radius is always perpendicular to tangent 

so       θ= 90⁰ ⇒   sin90= 1

So     

To calculate complete magnetic field, integrate both sides 

⇒     B =ʃ            

Here ʃ circumference of circle  

⇒    B =r

Or     B =

If the circular coil contains n turns then     

B =     

or    B= 

Here the direction of magnetic field can be given by right hand rule. 

Q4. The magnetic dipole moment of a current carrying coil does not depend upon:

A Number of turns of the coil      

B Cross-sectional area of the coil

C Current flowing in the coil                                 

D Material of the turns of the coil

Q5. An electron is revolving around the nucleus in a circular orbit with a speed of  m/s . If the radius of the orbit is  m, find the current constituted by the revolving electron in the orbit

Solution 

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.

Ans: 

Q7.  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?

Solution Since radius r = 10 cm = 0.1 m. The number of turns n = 100. 

So B= 

10 Magnetic field at a point on the axis of a circular coil carrying current:-

Suppose a point P on the axis of a Circular coil carrying current I. Suppose two small current element  &  on coil. Then the Magnetic field due to these current Elements at P is dB & dB’. As shown in fig. 

Clearly dB & dB’ are equal. Resolving dB & dB’ into rectangular components, we see that  &   are equal & opposite so they are cancelled out. 

So the total magnetic induction at P due to current through the whole circular coil is given  

B =ʃdB Sin∅            –(i)

From Ampere circuital law, the magnetic induction due to  at P is  

Here  is angle between r &  so 90⁰.      

So        dB =

Using in (i) 

B =   ʃ 

Here      sin∅       &   ʃ d =2

    ⇒    B =. 2

Or   

If there are n turn in the coil 

then   B = 

     at center  =0   ⇒       

B = 

     If P point lies far away from the coil then  a so a can be neglected 

⇒    B =

=

     If P point lies at a distance equal to radius of the coil I,e. r = a we have

⇒   B=  

= 

11 The direction of magnet field due to circular coil:- ^m.imp 

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.

 

 

 

 

 

 

12 Right hand rule:- 

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.

 

 

13 Clock rule:- 

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.

 

 

 

 

14 Ampere’s circuital law:- ^m.imp 

According to Ampere’s circuital law, the line integral of magnetic field around a closed circuit is equal to  times the total current flowing through the circuit.

I,e.             =I

Proof: –    

Suppose a Ampere’s surface around a straight conductor carrying current I. Now the magnet field at P due to current I in conductor can be given by    

Now the line integral of is   

  

=  

Here    = = circumference of the circle

So  

=  

Or       = 

This is required expression for Ampere circuital law.

     Ampere’s circular law holds for DC or steady current which does not change with time.

     Ampere’s circuital law for magnetic field is analogy to Gauss law in electrostatics.

     Biot Savart’s law & Ampere’s circuital law are equivalent in same sense as Coulombs Law & Gauss theorem in electrostatics.

     For a loop that is not connected to the current   

 

15 Applications of Ampere’s Circuital law ^m.imp 

  1. Magnetic field due to an infinite long current carrying wire:-

Suppose 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  at loop, then magnetic field B at  can be given by Ampere’s circuital law as 

   = 

O r  cos0⁰   =

 Or         B = 

Here  = circumference of circle =   

    B     =  

             B   =   

 

(b) Solenoid: – ^m.imp

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.

Calculation of magnetic field inside a long straight solenoid:-

Suppose a rectangular Amperean loop  of length. If N is the number of turns in the length, then total current through the loop equal to .

Thus according to Amperes law   

                            =  –(1)

Or       ʃ_abcda   =  ʃ_a^b +  ʃ_b^c +  ʃ_c^d + ʃ_d^a 

 Here      ʃ_b^c = ʃ_d^a =  ʃ cos  = 0

Also       ʃ_c^D   = 0 because CD lies outside the solenoid.

 = ʃ_a^b  = B               ..(2)

Comparing eqⁿ 1 & 2 we get       

 B = 

Or   B = 

Or    B =                

Where n is the number of turns per unit length.

Q8.How is the magnetic field inside a given solenoid made strong?

• By inserting a ferromagnetic substance inside the solenoid
• By increasing the amount of current through the solenoid

Q9. 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?

Solution   The number of turns per unit length is,  turns/m 

The length  and radius 

Thus, 

Hence, we can use the long solenoid formula, namely,

 B =   

 

c   Toroid or Toroidal Solenoid(not directly in cbse)

A solenoid bent into the form of a closed ring is called a toroidal solenoid.

Let us consider a toroid having N number of turn equally spaced & let I is the amount of current flowing through them.

•   Magnetic field at a point inside the core of the solenoid:-

From Ampere circuital law        

    =   

From N number of turns            

 = 

Here =                 

    B     = 

Or        B =   

Where n =  = number of turns per unit length, Thus magnetic is same as that of solenoid.

•    Magnetic field inside the toroid:-

 As inside a hollow conductor I = 0     

    

= 

    B=0

Thus Magnetic field inside the toroid is zero

•      Magnetic field outside the solenoid.

For any point outside the tortoid, the current threading the loop L’ through current at the point is zero so B = 0                                    

      = 0

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.      

          

16. Motion of a charge particle in a uniform electric field:-

Let us consider a charge particle + q moving with  velocity in the electric field then charge particle will experience a electric force.

As  = q-(1)

Due to this electric force charge particle starts accelerating with a force.

 = m-(2)

Comparing eqⁿ (1) & (2) we get

m = q

 = -(3)

Now let the time taken by the charge particle to move  distance is 

t = 

Similarly the distance travelled along y axis is

 = t + t²

Here = 0also = &

Soy=²

=

Or

wherek =

Ory  x²

Which is an eqⁿ of parabola, hence we can say that particle will follow parabolic path.

 

17. Motion of charge particle in uniform magnetic field:-

Let us suppose a charge particle +q enter in a magnetic field B at origin O with a velocity, making an angle θ with magnetic field. 

Now resolving  into two rectangular components we have ₁ = cosθ along B and ₂ =  sinθ along perpendicular to magnetic field, 

The force on the charge particle due to ₂ is 

Or

Or

The direction of this force can be obtained by R.H.R & is perpendicular to plane containing. 

Here motion of the charge particle is perpendicular to applied force so work done by this force is zero.

Hence this force provides no motion to the charge particle but gives the circular motion. i,e. centripetal force. So we may write

 =

or=

The angular velocity of the particle can be given as

The frequency of rotation of the particle in magnetic field will Becomes

 = 

=

So the time period

T = =

For component ₁ = cosθ the force is

=  = 0

Here there is no any force due to  so the charge particle moves linearly due to component.

Thus by the combined effect of & charge particle move linear as well as circular so its path becomes helical. The total distance travelled by the particle is 

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:

A.B.C.D.

Q11.An alpha particle is projected with velocityinto a region in which magnetic field exists. Calculate the acceleration of the particle in the region, charge to mass ratio for alpha particle is 

Ans:

Q12.What is the radius of the path of an electron (mass 9 × 10^-31 kg and charge 1.6 × 10^–19 C) moving at a speed of 3 ×10⁷ m/s in a magnetic field of 6 × 10^–4 T perpendicular to it?What is its frequency?Calculate its energy inkeV.( 1 eV = 1.6 × 10^–19 J).

SolutionAs

= 

= 26 ×  m = 26 cm

Hz=2 MHz.

 

18. Lorentz force ^m.imp

The total force experienced by a charge particle in electric & magnetic field both is called Lorentz force. 

The force experienced by charge particle in electric field is

=q–(1)

& the force experienced in magnetic field is

–(2)

Thus the total force (Lorenz force) experienced by charge particle is

=+ 

Or= q + 

OrF =q+

Special cases:-

1. When all ,  are collinear.  In this case  = 0 because

So the total force on the charge particle will be 

 = q

2. When, are mutually perpendiculars to each other, In this case    are such that

 =  + = 0, 

then  ==0

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.

 

19. Motion of charge particle in perpendicular magnetic & Electric field:- Or Velocity Selector:-

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. 

The electrons which have equal and opposite electric and magnetic field may pass through the velocity selector. 

The velocity of electrons in this case may be given as

 

20. Force on a current carrying conductor placed in magnetic field. ^m.imp

Consider a conductor PQ of length  and area. Suppose current flows in the conductor along + ve y axis & magnetic field is applied along +ve z axis. So the velocity  of electrons will be along-ve y axis. 

The magnetic force experienced by the electron may be given as

If there are n numbers of electrons per unit volume then total numbers of electrons are

So the total force experienced by the electrons is

(1)

As we know

Or

(As are in opposite direction so-ve sign is taken,) 

Now from e (1)

Or

where  is the direction between.

Direction of force:- 

The direction of force is given by Fleming’s left hand rule or Right hand thumb rule.

Special cases:-

1. If     θ  =  0°   or   180
2. then    sin0 = 0 F = 0

Hence conductor placed parallel to direction of magnetic field does not experience any force.

3. If =  90sin 90 = 1

When conductor is placed perpendicular to magnetic field it will experience maximum force.

Q13. The horizontal component of the earth’s magnetic field at a certain place is 3.0 ×10^–5T 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?

Solution 

The force per unit length is 

(a)When the current is flowing from east to west, θ = 90° 

Hence, 

This is larger than the value 2×10–7 Nm–1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s 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.

(b) When the current is flowing from south to north,  so hence there is no force on the conductor

 

21. Expression for the force between two Parallel current carrying wire^m.imp.

 

 

Consider two long parallel wires AB & CD carrying current. Let r be the separation between them. The magnetic field produced by current  at any point on wire CD is

This field acts perpendicular to the wire CD and points into the plane of the paper .It exerts a force given as

According to Fleming’s left hand rule, this force acts at right angle to CD, toward AB in the plane of the paper.

Similarly an equal force is exerted on the wire AB by the field of wire CD.

As 

Again according to Fleming’s left hand rule, this force acts at right angle to AB, toward CD in the plane of the paper. 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.

^Imp. The ampere is that constant current which—if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 meter apart in vacuum—would produce between these conductors a force equal toNewton’s per meter of length.

Q14.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. 

Solution 

()

 T

 

22. TExpression for the force between two Parallel current carrying wire:- ^m.imp

.Let us consider a current carrying rectangular coil of sides  such that area. 

Suppose this coil is placed at an angle θ with magnetic field B. 

Now as we know that when a current carrying conductor placed in magnetic field experience a force, 

So force on side PQ

() 

= IlBsin90⁰ = I…..(1)

Again Force on side QR

= () 

 = ………..…..(2)

And force on side RS

()

=

= I…….(3)

Similarly Force on side SP

()

 = 

 =  …….(4)

According to Fleming’s left hand rule the Forces  &  are equal and act along the axis of the coil in opposite direction so exert no any torque they are cancelled out. 

While the force  &  will exert a torque on the coil given as

()

0r

Where = magnetic dipole moment 

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

1. Total torque on the coil, 

2. Total force on the coil. 

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  and the free electron density is .

Ans: 

1. 
2. 
3. 

 

23Moving coil Galvanometer:- ^m.imp

A Galvanometer is a device which is used to detect the current in a circuit.

Principle:-

The Galvanometer is based on the principal that a current carrying coil placed in external magnetic field experience a torque.

Construction:- –

A moving coil Galvanometer consist of

1.             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’. An arrangement of strong north & South Pole is fixed on both sides of the coil.
2.             A soft iron core L of spherical or cylindrical shape is placed between the coils.
3.              A circular Mirror M is attached to the Phosphor bronze strip, to measure the deflection of the coil.
4.              The torsion head is connected to a binding terminal T₁ & spring S’ is connected to another binding terminal T₂.

Theory:-

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 

Here θ is angle between normal to the plane of coil & applied magnetic field.

Since the field is radial

i,eθ = 90⁰ 

Due to this torque the phosphor bronze wire suffer twist, if θ is angle of twist & k is moment of restoring couple per unit angular twist then moment of restoring couple = k θ

In equilibrium

Or

Here  = K = a another constant 

SoI  θ

Hence current through the coil is directly proportional to the deflection of the coil.

 

Sensitivity of a moving coil Galvanometer

A galvanometer is said to be more sensitive if it show large deflection even for a small change in current is passed through it.

1. Current sensitivity: – it is defined as the deflection produced in the galvanometer when a unit current is passed through it.

I,e. Current sensitivity

2. Voltage sensitivity: – It is defined as the deflection produced in the galvanometer when a unit potential difference is applied across it.

Voltage sensitivity V_s==

=

Clearly voltage sensitivity =

The sensitivity of a moving coil galvanometer can be increased by 

By increasing number of turns in the coil.

By increasing the magnetic field.

By increasing the area of the oil.

By decreasing the value of torsion constant R

Advantage of moving coil Galvanometer:-

• It tells about the deflection due to current in a circuit.

Disadvantage of a moving coil Galvanometer:

• Its sensitivity cannot be changed at will.
• It is damaged by overloading. 

 

Q16.What is the importance of a radial magnetic field and how is it produced?

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.

24 Conversion of a Galvanometer into an ammeter ^m.imp:-

(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. 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.)

As shunt & Galvanometer are connected in parallel 

SoG = (I – ) S

Or

So by connecting a shunt across galvanometer, we get an ammeter of desired range, 

Moreover

 = 

As the deflection in galvanometer is directly proportional to  & hence perpendicular so the current can be measured. Also the effective resistance is

R –  < S

• An ideal ammeter has zero resistance.(imp:-MCQ)
• Higher the range of ammeter to be prepared lower should be the value of shunt.
• The range of ammeter can be increased but cannot be decreased.

Q17. A galvanometer has a resistance of 30Ω. 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.

Ans: As 

Substitution so 

25 Conversion of a galvanometer into a Voltmeter:- ^m.imp

(Voltmeter is a device which is used to measure the potential difference between two points in a circuit so connected in parallel. A voltmeter is formed by connecting a high resistance in series a Galvanometer.)

The total resistance in the Circuit is 

Now by ohm’s Law

 = 

Or

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

Its effective resistance is 

• An ideal voltmeter should have infinite resistance.(imp:-MCQ)

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?

Reason:

Voltmeter: Thisensuresthataverylowcurrentpassesthroughthevoltmeterandhencedoesnotchange(much)the original potential difference to be measured.

Ammeter: This ensures that the total resistance of the circuit does not change much and the current flowing remains (almost) at its original value.

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 ; (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance   r s = 0.02Ω; (c) is an ideal ammeter with zero resistance?

Solution (a) Total resistance in the circuit is .

Hence

1.        Resistance of the galvanometer converted to an ammeter is,

Total resistance in the circuit is, 

Hence,

2. For the ideal ammeter with zero resistance,