Answer to Question #104218 in Electricity and Magnetism for Akshay

Question

A wire loop of resistance
10 ohm and radius 10 cm is kept in the plane of this paper in a uniform magnetic field B.
The direction of B is perpendicular to the
plane of the page and points out of it and its magnitude is increasing at the rate of 0.50T/s.
Determine the magnitude and direction of the induced current in the loop.

Answer 1

From the law of electromagnetic induction we write

$\epsilon=\frac{\Delta\Phi}{\Delta t}=\frac{\Delta(B \cdot S)}{\Delta t}=(\frac{\Delta B }{\Delta t})\cdot S$

Where

$S=\pi \cdot r^2$

$(\frac{\Delta B }{\Delta t})$ -The rate of change of the magnetic field

Then the current is

$I=\frac{\epsilon}{R}=\frac{(\frac{\Delta B }{\Delta t})\cdot S}{R}=\frac{(\frac{\Delta B }{\Delta t})\cdot \pi \cdot r^2}{R}=\frac{(0.5)\cdot 3.14 \cdot 0.1^2}{10}=1.57 mA$