Ampere’s circuital law

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.

Chapter 4 Moving Charges and Magnetism

class 12 physics notes