Q: A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. The magnitude of current in this time interval is.

Sol: The initial magnetic flux is given by

φ = BA cosθ

Given, B = 0.10 T, area of square loop = 10 × 10 = 100 cm^{2} = 10^{-2} m^{2}

$\large \phi = \frac{0.1 \times 10^{-2}}{\sqrt{2}} Wb$

Final Flux = 0

The change in flux is brought about in 0.70 s the magnitude of the induced emf is

$\large e = \frac{\Delta \phi}{\Delta t} = \frac{\phi – 0}{\Delta t} $

$\large = \frac{10^{-3}}{\sqrt{2} \times 0.7} = 1 mV $

The magnitude of current is $\large I = \frac{e}{R} = \frac{10^{-3}}{0.5} $

I = 2 m A