Question

Suppose that the radius of cross-section of the wire used in the figure is $2\times {10}^{-3}\text{unit}$. The increase in the radius of the loop is $\frac{I{a}^{2}B}{\pi NY}$ if the magnetic field $B$ is switched off. The Young's modulus of the material of the wire is $Y$. Find the value of $N$.

• A. $\frac{{10}^6}{4}$
• B. $\frac{{10}^6}{2}$
• C. $2\times {10}^{-6}$
• D. $4\times {10}^{-6}$

Answer 1

The correct option is D $4\times {10}^{-6}$

Using,
$Y=\frac{\text{stress}}{\text{strain}}=\frac{\frac{F}{\pi r^2}}{\frac{dl}{L}}$
Where,

$F=IaB\to$ Compressive magnetic force,

$r\to$ Radius of cross-section of wire,

$dl\to$ Change in length of loop,

$L=2\pi a\to$ Length of the loop,

$\Rightarrow \frac{dl}{L}Y=\frac{F}{\pi r^2}\Rightarrow dl=\frac{F}{\pi r^2}\times \frac{L}{Y}$

$\Rightarrow dl=\frac{2\pi a^{2}IB}{\pi r^{2}Y}$

For small cross-sectional circle, $dl=2\pi da$, where, $da$ is the change in radius of the loop.

$\Rightarrow da=\frac{2\pi a^{2}IB}{\pi r^{2}Y}\times \frac{1}{2\pi }=\frac{a^{2}IB}{\pi r^{2}Y}$

Given, $r=2\times {10}^{-3}\text{unit}$

$\Rightarrow da=\frac{a^{2}IB}{\pi \times (2\times {10}^{-3}{)}^2\times Y}=\frac{a^{2}IB}{4\times {10}^{-6}\pi Y}$

According to the problem,

$da=\frac{a^{2}IB}{\pi NY}$

$\therefore N=4\times {10}^{-6}$

Hence, option $\left(D\right)$ is the correct answer.