Question

The correct relation between B and M for a small current carrying coil is:
A) $B = \dfrac{{{\mu _0}M}}{{2{x^3}}}$
B) $B = \dfrac{{{\mu _0}M}}{{{x^3}}}$
C) $B = \dfrac{{{\mu _0}M}}{{\pi {x^3}}}$
D) $B = \dfrac{{{\mu _0}M}}{{2\pi {x^3}}}$

Answer 1

Hint A small amount of current is flowing in a coil. We have to derive a relation between B and M. We will use definitions of B and M to find which one is a correct formula satisfying the relation.

Complete step by step solution
Both B and M are vector quantities.
B - It is magnetic flux density. The amount of magnetic field lines crossing a unit area normally from a magnetic material. Strength of the magnetic field is given by this quantity.
It's S.I. unit is \[\dfrac{{kg}}{{{s^2}}}/A\] or Tesla.
 $B = \dfrac{{{\mu _0}I}}{{2x}}$ …(1)
Here x= radius of circle and ${\mu _0}$ is magnetic susceptibility.
Value of ${\mu _0}$ is $4\pi \times {10^{ - 7}}\dfrac{{Tm}}{A}$ .
M -It is a magnetization field. It is also known as magnetic polarization. When current is passed from all the dipoles aligns itself in a particular direction. This sets up magnetic moments which get induced in the substance. The response of a magnet on the application of magnetic field gives resultant magnetization. Hence it is also explained as magnetic moment per unit volume.
 $M = \dfrac{m}{V} = IA$ …(2)
 \[
  I = \dfrac{M}{A} \\
  A = \pi {x^2} \\
 \]
Using value of equation (2) in (1)
 $B = \dfrac{{{\mu _0}M}}{{2\pi {x^3}}}$

Hence the correct option is (D).

Note
If $B = \dfrac{{{\mu _0}I}}{x}$ is used then we get $B = \dfrac{{{\mu _0}M}}{{\pi {x^3}}}$ I.e. option (C) as our solution. But this is not the correct option. If we have used $M = \dfrac{m}{V}$ then we might not reach the solution. So, we cannot use this value. Option A and B are also wrong. Therefore, we are left with only one option I.e. D which is correct.