Question

A conducting ring of radius $1\, m$ is placed in an uniform magnetic field $B$ of $0.01\, T$ oscillating with frequency $100\, Hz$ with its plane at right angle to $B$. What will be the induced electric field?

• A. $\pi$ volts/m
• B. 2 volts/m
• C. 10 volts/m
• D. 62 volts/m

Answer 1

The Correct Option is B

From Faraday's law of electromagnetic induction the induced emf is equal to negative rate of change of magnetic flux. That is $e=-\frac{\Delta \phi}{\Delta t}$ Flux induced $=2 B A \cos \phi$ where $B$ is magnetic field, $A$ is area. Given, $\theta=0^{\circ}, \Delta t=\frac{1}{100} s$ $\Delta \phi=2 \times 0.01 \times \pi \times(1)^{2} \times 200 \times \cos 0^{\circ}$ $\therefore \quad e=\frac{-2 \times 0.01 \times \pi \times(1)^{2} \times 200}{100}$ $=-4 \pi$ Volt Circumference of a circle of radius $r$ is $2 \pi r$. $\therefore$ Induced electric field $E$ is $E=\frac{|e|}{2 \pi r}=\frac{4 \pi}{2 \pi r}=\frac{2}{1}=2\, V / m$