Question

A single-turn plane loop of wire with a cross-sectional area 200 cm2 is perpendicular to a magnetic field that increases uniformly from 0.200 T to 2.800 T in 2.20 seconds. What is the magnitude of the induced current if the resistance of the coil is 8.00 Ω?

Answer 1

Answer:

Induced current,$I=2.87\times 10^{-3}\ A$                                        

Explanation:

Given that,

Area of cross section of the wire, $A=200\ cm^2=0.02\ m^2$

Time, t = 2.2 s

Initial magnetic field, $B_i=0.2\ T$

Final magnetic field, $B_f=2.8\ T$

Resistance of the coil, R = 8 ohms

The expression for the induced emf is given by :

$\epsilon=-\dfrac{d\phi}{dt}$

$\phi$ = magnetic flux

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=A\dfrac{d(B)}{dt}$

$\epsilon=A\dfrac{B_f-B_i}{t}$

$\epsilon=0.02\times \dfrac{2.8-0.2}{2.2}$

$\epsilon=-0.023\ volts$

So, the induced emf in the loop is 0.023 volts. The induced current can be calculated using Ohm's law as :

$\epsilon= IR$

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{0.023}{8}$

$I=2.87\times 10^{-3}\ A$

So, the magnitude of the induced current in the loop of wire is $2.87\times 10^{-3}\ A$. Hence, this is the required solution.