Question

A magnetic field $B = B_0 \sin \omega t$ where $t$ is time, is applied perpendicular to the plane of a square coil of side $a$ and resistance $R$. Net electric charge that flows through the coil during time interval $t = 0$ to $t = \frac{2\pi}{\omega}$ is

• A. $\frac{B_0 a}{R^2}$
• B. $\frac{B_0 a^2}{R}$
• C. Zero
• D. $\frac{2B_0 a^2}{R}$

Answer 1

Answer: (C)

Zero

Answer 2

The magnetic flux through the coil is

$\Phi = B_0 \sin (\omega t) \cdot a^2$.

The induced emf is

$\mathcal{E} = -\frac{d\Phi}{dt} = -a^2 B_0 \omega \cos (\omega t)$.

Then, the current in the coil is

$I = \frac{\mathcal{E}}{R} = -\frac{a^2 B_0 \omega \cos (\omega t)}{R}$.

The net charge $Q$ that flows is obtained by integrating the current over one complete cycle (from $t=0$ to $t=\frac{2\pi}{\omega}$):

$Q = \int_0^{\frac{2\pi}{\omega}} I\, dt = -\frac{a^2 B_0 \omega}{R} \int_0^{\frac{2\pi}{\omega}} \cos (\omega t)\, dt$.

Making the substitution $u=\omega t$ (so that $du = \omega dt$), the integral becomes:

$\int_0^{2\pi} \cos u\, \frac{du}{\omega} = \frac{1}{\omega} \left[\sin u\right]_0^{2\pi} = \frac{1}{\omega}(0) = 0$.

Thus,

$Q = 0$.