Question

The axial magnetic fields due to infinite solenoid

Answer 1

Answer:

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

Let OP=r

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=μ0ia2ndx2[(r−x)2+a2]3/2

If r>>a and r>>x then

dB=μ0ia2ndx2r3

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

∫dB=μ0ina22r3∫−lldx

B=μ0nia22r3(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)

B=μ04π2Mr3

This is the expression for magnetic field on the axial field of finite solenoid carrying current is same as that of a bar magnet .