A planar loop consisting of four turns of wire,each of which encloses 200 cm²,is oriented perpendicularly to a magnetic field that increases uniformly in magnitude from 10 mT to 25 mT in a time of 5.0 ms.What is the resulting induced current in the coil if the resistance of the coil is 5.0 W? A)60 mA B)12 mA C)0.24 mA D)48 mA

Question

Quizplus · Accepted Answer

The magnetic flux through each turn of the loop is given by $\phi = BA$, where $B$ is the magnetic field and $A$ is the area enclosed by each turn. Since there are four turns, the total flux through the loop is $4\phi = 4BA$. At $t=0$, the flux through each turn is $\phi_0 = 10^{-2} 	ext{ T} \cdot 0.02 	ext{ m}^2 = 2	imes 10^{-4} 	ext{ Wb}$. At $t=5.0	ext{ ms}$, the flux through each turn is $\phi_1 = 25 	imes 10^{-3}	ext{ T} \cdot 0.02 	ext{ m}^2 = 5 	imes 10^{-4} 	ext{ Wb}$. Therefore, the change in flux through each turn is $\Delta \phi = \phi_1 - \phi_0 = 3	imes 10^{-4} 	ext{ Wb}$. The average emf induced in the loop over the time interval $\Delta t = 5.0	ext{ ms}$ is
$$
\bar{\epsilon} = \frac{\Delta \phi}{\Delta t} = 60 	ext{ V}.
$$
The induced current is given by Ohm's law: $I = \bar{\epsilon}/R = 12	ext{ mA}$. Therefore, the best choice is (D).

A Planar Loop Consisting of Four Turns of Wire,each of Which