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 $4mm$ and a total length of $30cm$. The magnetic field changes with time at a steady rate $dB/dt=0.032T{s}^{-1}$. The induced current in the loop is close to (Resistivity of the metal wire is $1.23\times {10}^{-8}\Omega m$

• A. $0.53A$
• B. $0.61A$
• C. $0.34A$
• D. $0.43A$

Answer 1

The correct option is B

$0.61A$

Step 1. Given data and draw a diagram:

Diameter, $d=4mm=0.004m$

Length, $l=30cm=0.3m$

Rate of change of magnetic field,$dB/dt=0.032T{s}^{-1}$

Resistivity of metal, $\rho =1.23\times {10}^{-8}\Omega m$

Step 2. Finding resistance, $R$ :

Area of cross-section of wire, $A_w=\pi r^2$

$R=\frac{\rho l}{A_w}$ $=\frac{1.28\times {10}^{-8}\times 0.3}{3.14\times {(0.002)}^2}$

$=3.06\times {10}^{-4}\Omega$

Step 3. Finding emf induced due to magnetic field :

Length of the side of a square, $a=\frac{l}{4}$

Area of the square loop, $A_l=\frac{l^2}{16}$

Magnetic flux, $\varphi =B{A}_l$

$emf$ induced is, $e=\frac{d\varphi }{dt}=\frac{dB}{dt}{A}_l=\frac{dB}{dt}\frac{l^2}{16}$

$e=0.032\times \frac{{(0.3)}^2}{16}$

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

Step 4. The induced current is :

$i=\frac{e}{R}=\frac{1.8\times {10}^{-4}}{3.06\times {10}^{-4}}$

$=0.58A\approx 0.6A$

Hence, option $B$ is correct.