Question

A uniform magnetic field $B$ exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of $4\text{mm}$ and a total length of $30\text{cm}$. The magnetic field changes with time at a steady rate $\frac{dB}{dt}=0.032{\text{Ts}}^{-1}.$ The induced current in the loop is close to (Resistivity of the metal wire is $\rho =1.23\times {10}^{-8}\Omega \text{m}$)

• A. $0.43\text{A}$
• B. $0.61\text{A}$
• C. $0.34\text{A}$
• D. $0.53\text{A}$

Answer 1

The correct option is B $0.61\text{A}$

Given,
$\text{Length of wire},l=30\text{cm}$
$\text{Radius of wire,}r=2\text{mm}$

Emf generated, $|E|=\frac{d\varphi }{dt}=\frac{dB}{dt}\left(A\right)(\because \varphi =B.A.)$

Here, $A\to \text{Area of the loop}$

As, the total length of the square loop is $30\text{cm}$, side of the square loop will be,

$a=\frac{30}{4}=7.5\text{cm}$

$\Rightarrow A=7.5\times 7.5{\text{cm}}^2=56.25\times {10}^{-4}{\text{m}}^2$

$\Rightarrow E=0.032\times 56.25\times {10}^{-4}$

$\Rightarrow E=1.8\times {10}^{-4}\text{V}$

Current, $i=\frac{E}{R}$

But, resistance of wire, $R=\rho \frac{l}{A^{\prime }}$

Here, $A^{\prime }$ is the area of cross-section of the wire.

$\Rightarrow A^{\prime }=\pi {(2\times {10}^{-3})}^2$

$\Rightarrow A^{\prime }=12.56\times {10}^{-6}{\text{m}}^2$

$\Rightarrow R=\frac{1.23\times {10}^{-8}\times 0.3}{12.56\times {10}^{-6}}$

$\Rightarrow R=2.93\times {10}^{-4}\Omega$

$\Rightarrow i=\frac{1.8\times {10}^{-4}}{2.93\times {10}^{-4}}$

$\Rightarrow i=0.61\text{A}.$

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