Question

A square loop of side $12 \,cm$ and resistance $0.60 \, \Omega$ is placed vertically in the east-west plane. A uniform magnetic field of $0.10\, T$ is set up across the plane in north-east direction. The magnetic field is decreased to zero in $0.6 \,s$ at a steady rate. The magnitude of current during this time interval is

• A. $1.42 \times 10^{-3}\,A$
• B. $2.67 \times 10^{-3}\,A$
• C. $3.41 \times 10^{-3}\,A$
• D. $4.21 \times 10^{-3}\,A$

Answer 1

The Correct Option is B

Here, Area $A=l^{2}=(12\, cm)^{2}$ $=1.4\times 10^{-2}\,m^{2}$ $R=0.60\, \Omega$, $ B_{1}=0.10\, T$, $\theta=45^{\circ}$ $B_{2}=0$, $dt=0.6\, s$ Initial flux, $\phi_{1}=B_{1}A\, cos\, \theta$ $=0.10 \times1.4\times10^{-2}\times cos\, 45^{\circ}$ $= 9.8 \times 10^{-4}\, Wb$ Final flux, $\phi_{2}=0$ Induced emf, $\varepsilon=\frac{\left|d\phi\right|}{dt}=\frac{\left|\phi_{2}-\phi_{1}\right|}{dt}$ $=\frac{\left|9.8\times10^{-4}\right|}{0.6\,s}=1.6\times10^{-3}\,V $ Current, $I=\frac{\varepsilon}{R}=\frac{1.6\times10^{-3}}{0.6}$ $2.67\times10^{-3}\,A$