Question

A very small circular loop of area $5\times {10}^{-4}{\text{m}}^2$ of resistance $2\Omega$ and negligible inductance is initially coplanar and concentric with a much larger fixed circular loop of radius $0.1\text{m}$. A constant current of $1\text{A}$ is passed in the bigger loop and the smaller loop is rotated with angular velocity $1\text{rad/s}$ about a diameter. Calculate the maximum flux linked with the smaller loop.

• A. $2\pi \times {10}^{-9}\text{Wb}$
• B. $4\pi \times {10}^{-9}\text{Wb}$
• C. $\pi \times {10}^{-9}\text{Wb}$
• D. $\frac{\pi }{2}\times {10}^{-9}\text{Wb}$

Answer 1

The correct option is C $\pi \times {10}^{-9}\text{Wb}$

The magnetic field due to the large loop at the centre is,

$B=\frac{{\mu }_{0}i}{2a}$

From data given in the question,

$B=\frac{{\mu }_0\times 1}{2\times 0.1}=5{\mu }_0$

As the size of the smaller loop is very less, we can consider magnetic field to be uniform throughout the loop.

Flux linked with the small loop,

$\varphi =BA\cos \theta$

$=5{\mu }_0\times 5\times {10}^{-4}\cos \theta$

$=2.5\times {10}^{-3}{\mu }_0\cos \theta$

Flux will be maximum for $\theta =0^{\circ }$.

$\therefore {\varphi }_{max}={\mu }_0\times 2.5\times {10}^{-3}\times \cos 0^{\circ }$

$=4\pi \times {10}^{-7}\times 2.5\times {10}^{-3}\times 1$

$=\pi \times {10}^{-9}\text{Wb}$

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