Question

State and prove ampere's law?

Answer 1

Ampere’s circuital law states that line integral of magnetic field forming a closed loop around the current(i) carrying wire, in the plane normal to the current, is equal to the μo times the net current passing through the close loop.



Here μo = permeability of free space = 4π×10-15N/A2

This law is based on the assumption that the closed loop consists of small elemental parts of length dl, and the total magnetic field of the closed loop will be the integral of magnetic field and the length of these elements This closed loop is called Amperian loop

Further, this integral will be equal to the multiplication of net current passing through this closed loop and the permeability of free space(μoi)

Proof-1(Regular coil):



To prove:        ∫B.dl = μoi

Starting from the left hand side, we can see in the diagram that angle between the element dl and magnetic field B is 0°



 We know that magnetic field due to a long current carrying wire is:

B = μoi/(2πr)

Also, the integral of element will form the whole circle of circumference (2πr):

∫ dl = 2πr

Now putting the value of B and ∫ dl in the equation, we get:

B∫ dl = μoi/(2πr) × 2πr = μoi

∴∫B.dl = μoi

Answer 2

Ampere's law states that “The magnetic field created by an electric current is proportional to the size of that electric current with a constant of proportionality equal to the permeability of free space.”

The line integral of the magnetic field B around any closed path is equal to μ0 times the net steady current enclosed by this path. Proof: The magnetic field produced by a long straight conductor is in the form of concentric circles. These circles are in the plane perpendicular to the length of the conductor.

oint overline B . overline dl = mu o NI

But overline B and overline dl are in the same direction.

. overline B overline dl =B.dl.Cos 0^ o =B.dl

.. oint overline B overline dl = oint B.dl=B. oint dl

.. § overline B . overline dl =B.(2 pi r) ….….….….….….…....... (2)

From equations (1) and (2)

B.(2 pi r)= mu o NI

B = (mu_{o}*NI)/(2pi*r) (3)

If 'n' is number of turns per unit length then n= - 2

Substituting this value in equation(3) we get

B = mu_{0}*nI

I don't know it's correct answer but this might help .

#SPJ2