Question

Obtain an expression for axial magnetic field of a current loop in terms of its magnetic moment.

Answer 1

Let the radius of the solenoid = a

length of the solenoid whose centre is o = 2l

No of turns per units length of solenoid = n

strength of current passed through the solenoid = i  

Consider a small element dx of the solenoid at a distance x from 0.

number of turns in the element = ndx

Hence the magnitude of magnetic field at P due to this element is

dB=μ0ia2ndx/2[(r−x)2+a2]3/2

If r>>a and r>>x then

dB=μ0ia2ndx/2r3

The range of variation of x is from x= −l to x= +l

∫dB=μ0ina22r3 ∫dx

B=μ0nia2 /2r3(2l)

M is the magnetic moment of the solenoid , then

M= total number of turns ×× current ×× area of cross section

M=n(2l)×i(πa2)M=n(2l)×i(πa2)

B=(μ0/4π)(2M/r3)

This is the expression for magnetic field on the axial field of finite solenoid carrying current.