Question

A circular coil of radius 1.5 cm carries a current of 1.5 A.If the coil has 25 turns ,the magnetic field at the centre of the coil is found to be M x 10${}^{-3}$T.What is the value of M ?(considering $\pi$ = 3.14)

Answer 1

Radius of each turn, $r=1.5cm$

Current flow in the coil, $I=1.5A$

Magnitude of magnetic field at the center $\left(B\right)$ is given by

$\left|B\right|=\frac{{\mu }_{\circ }}{4\pi }\frac{2\pi nI}{r}$

where $\mu$ is permeability of free space.

${\mu }_{\circ }=4\pi \times {10}^{-7}Tm{A}^{-1}$

$\left|B\right|=\frac{4\pi \times {10}^{-7}}{4\pi }\times \frac{2\times \pi \times 25\times 1.5}{1.57\times {10}^{-2}}$

$\therefore M\times {10}^{-7}=1.57\times {10}^{-3}$

so, we get

$M=1.57$