A circular loop,of radius 15 cm and negligible resistance,is sitting in a perpendicular magnetic field of 0.1 T.If the magnetic field strength changes to a value of 0.5 T in 0.5 s,calculate the induced current in the loop if it is in series with a 20 ohm resistor. A)5.7 $\times$ 10^-2 A B)1.2 $\times$ 10^-2 A C)2.8 $\times$ 10^-3 A D)6.0 $\times$ 10^-4 A E)1.1 A

Question

Quizplus · Accepted Answer

The induced emf in the loop is given by $\epsilon=-\frac{d\phi}{dt}$, where $\phi$ is the magnetic flux through the loop. 
When the magnetic field changes from 0.1 T to 0.5 T, the change in flux through the loop is given by $\Delta\phi=BA\cos	heta$, where $A$ is the area of the loop and $	heta$ is the angle between the magnetic field and the normal to the loop. Since the loop is perpendicular to the magnetic field, $	heta=0$ and $\Delta\phi=BA=0.15^2\pi(0.5-0.1)=0.0141$ Wb.
The time taken for this change in flux is $\Delta t=0.5$ s. Therefore, the induced emf is $\epsilon=-\frac{d\phi}{dt}=-\frac{\Delta\phi}{\Delta t}=-0.0282$ V.
The induced current in the loop is given by $I=\frac{\epsilon}{R}$, where $R$ is the resistance of the loop and the resistor. Since the loop has negligible resistance compared to the 20 ohm resistor, we can assume that all the voltage drops across the resistor. Therefore, $R=20$ ohms.
Substituting the values, we get $I=\frac{-0.0282}{20}=-0.00141$ A or $1.41	imes10^{-3}$ A. 
Therefore, the best choice is (C) 2.8 mA.

A Circular Loop,of Radius 15 Cm and Negligible Resistance,is Sitting ×\times×

A Circular Loop,of Radius 15 Cm and Negligible Resistance,is Sitting $\times$