Question

A uniform magnetic field $B$ exists in a cylindrical region of radius $10\text{cm}$ as shown in the figure. A uniform wire of length $80\text{cm}$ and resistance $4.0\Omega$ is bent into a square frame and is placed with one side along a diameter of the cylindrical region. If the magnetic field increases at a constant rate of $0.010\text{T/s}$, find the current induced in the frame.

• A. $3.9\times {10}^{-5}\text{A}$
• B. $3.9\times {10}^{-4}\text{A}$
• C. $1.5\times {10}^{-4}\text{A}$
• D. $1.7\times {10}^{-5}\text{A}$

Answer 1

The correct option is A $3.9\times {10}^{-5}\text{A}$

Here, the flux due to the magnetic field $B$ only passes through the semicircular area of the frame.

Therefore, the flux through the wire frame is,

$\varphi =BA\cos \theta =B\times \frac{\pi R^2}{2}\times \cos 0^{\circ }=\frac{\pi R^{2}B}{2}$

Now, magnitude of the emf induced in the frame is given by,

$ϵ=\frac{d\varphi }{dt}=\frac{d}{dt}\left(\frac{\pi R^{2}B}{2}\right)$

$\Rightarrow ϵ=\frac{\pi R^2}{2}\frac{dB}{dt}$

From the data given in the question,

$ϵ=\frac{\pi \times {0.1}^2}{2}\times 0.01$

$\Rightarrow ϵ=1.57\times {10}^{-4}\text{V}$

$\therefore$ Current in the frame,

$i=\frac{ϵ}{R}=\frac{1.57\times {10}^{-4}}{4}=3.9\times {10}^{-5}\text{A}$