Question

A circular loop of radius 20 cm and resistance 2 \(\Omega\) is placed in a time varying magnetic field \( \vec{B} = (2t^2 + 2t + 3) \text{ T} \). At \( t=0 \), for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at \( t = 3 \text{ s} \) is \( \alpha/50 \text{ A} \). The value of \( \alpha \) is \dots (Take \( \pi = 22/7 \))}

Answer 1

Correct Answer: 44

Step 1: Understanding the Concept:
According to Faraday's law of induction, a change in magnetic flux through a circuit induces an electromotive force (emf). The induced current can then be found using Ohm's law.
: Key Formula or Approach:
Flux \( \Phi = B \cdot A \cdot \cos \theta \) (here \( \theta = 0^\circ \)).
Induced emf \( e = \left| \frac{d\Phi}{dt} \right| = A \cdot \left| \frac{dB}{dt} \right| \).
Current \( I = e/R \).
Step 2: Detailed Explanation:
Radius \( r = 20 \text{ cm} = 0.2 \text{ m} \).
Area \( A = \pi r^2 = \frac{22}{7} \cdot (0.2)^2 = \frac{22 \cdot 0.04}{7} = \frac{0.88}{7} \text{ m}^2 \).
Magnetic field \( B = 2t^2 + 2t + 3 \).
Rate of change \( \frac{dB}{dt} = 4t + 2 \).
At \( t = 3 \text{ s} \), \( \frac{dB}{dt} = 4(3) + 2 = 14 \text{ T/s} \).
Induced emf:
\[ e = A \cdot \frac{dB}{dt} = \frac{0.88}{7} \times 14 = 0.88 \times 2 = 1.76 \text{ V} \].
Induced current:
\[ I = \frac{e}{R} = \frac{1.76}{2} = 0.88 \text{ A} \].
Given \( I = \frac{\alpha}{50} \implies 0.88 = \frac{\alpha}{50} \).
\[ \alpha = 0.88 \times 50 = 44 \].
Step 3: Final Answer:
The value of \( \alpha \) is 44.