Question

A square loop of each side 10 cm and having resistance 0.5 $\Omega$ is kept vertical in east-west plane. The uniform magnetic flux density of 0.01 Tesla is established across the plane along north-east direction. After 0.70 sec, the magnetic field is reduced to zero with uniform rate. Calculate the following:
(i) the initial and final magnetic flux,
(ii) the induced e.m.f.,
(iii) the induced current,
(iv) the induced charge.

Hint:

Always use Faraday’s law: $\mathcal{E} = -\dfrac{d\Phi}{dt}$ and Ohm’s law for induced current.

Answer 1

Step 1: Given data.
Side of square loop $a = 0.10 \, m$, area $A = a^2 = 0.01 \, m^2$
Resistance $R = 0.5 \, \Omega$
Magnetic field $B = 0.01 \, T$
Angle between $B$ and area vector = $45^\circ$ (since along north-east).
Step 2: Initial and final flux.
Flux $\Phi = B A \cos\theta$
\[ \Phi_i = 0.01 \times 0.01 \times \cos 45^\circ = 1 \times 10^{-4} \, Wb \] \[ \Phi_f = 0 \]
Step 3: Induced emf.
Faraday’s law: \[ \mathcal{E} = \frac{\Delta \Phi}{\Delta t} = \frac{1 \times 10^{-4}}{0.70} \approx 1.43 \times 10^{-4} \, V \]
Step 4: Induced current.
\[ I = \frac{\mathcal{E}}{R} = \frac{1.43 \times 10^{-4}}{0.5} \approx 2.86 \times 10^{-4} \, A \]
Step 5: Induced charge.
\[ Q = I \times t = 2.86 \times 10^{-4} \times 0.70 \approx 2.0 \times 10^{-4} \, C \]
Step 6: Conclusion.
- $\Phi_i = 1 \times 10^{-4} \, Wb$, $\Phi_f = 0$
- $\mathcal{E} = 1.43 \times 10^{-4} \, V$
- $I = 2.86 \times 10^{-4} \, A$
- $Q = 2.0 \times 10^{-4} \, C$