Question

A long straight solenoid of cross-sectional diameter $d$ and with $n$ turns per unit of its length has a round loop of copper wire of cross-sectional area $A$ and resistivity $\rho$. The loop is lying at the centre of the solenoid, with its axis parallel to the axis of the solenoid. Find the current flowing in the loop if the current in the solenoid winding is increased with rate $I\text{Ampere/sec}$. The diameter of the loop is equal to the diameter of the solenoid.

• A. $\frac{{\mu }_{0}ndAI}{\rho }$
• B. $\frac{2{\mu }_{0}ndAI}{\rho }$
• C. $\frac{{\mu }_{0}ndAI}{2\rho }$
• D. $\frac{{\mu }_{0}ndAI}{4\rho }$

Answer 1

The correct option is D $\frac{{\mu }_{0}ndAI}{4\rho }$

The magnetic field inside the solenoid is given as,

$B={\mu }_{0}ni$

The flux through in cross-section of copper winding is given as,

$\varphi =BA=\left({\mu }_{0}ni\right)\frac{\pi d^2}{4}$

The $\text{emf}$ induced in the copper wire is given as,

$E=\left|\begin{array}{c}-\frac{d\varphi }{dt}\end{array}\right|={\mu }_{0}n\pi \frac{d^2}{4}\frac{di}{dt}={\mu }_{0}n\pi \frac{d^2}{4}I[\text{Given,}\frac{di}{dt}=I]$

So, the current induced in the copper wire will be,

$i_{ind}=\frac{E}{R}=\frac{\frac{{\mu }_{0}n\pi d^{2}I}{4}}{\rho \left(\frac{\pi d}{A}\right)}=\frac{{\mu }_{0}ndAI}{4\rho }$

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