Question

A coil of one loop is made from a wire of length L and there after a coil of two loops is made from same wire, then the ratio of magnetic field at the centre of coils will be

• A. 1:4
• B. 1:1
• C. 1:8
• D. 4:1

Answer 1

The correct option is A 1:4

Given : Length of the wire is $L$

Let radius of first and second loop be $R_1$ and $R_2$ respectively.

According to the question $L=2\pi R$ (first loop) ..............(1)

Also $L=2\times 2\pi R_2$ (second loop) .............(2)

From (1) and (2), $2\pi R_1=2\left(2\pi R_2\right)$ $⟹\frac{R_2}{R_1}=\frac{1}{2}$

Magnetic field at the center of the coil of radius $R$ is given by $B=\frac{{\mu }_{o}nI}{2R}$

$⟹$ $\frac{B_1}{B_2}=\frac{n_1{R}_2}{n_2{R}_1}=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$ $(\because n_2=2,$ $n_1=1)$

Thus $B_1:B_2=1:4$