Question

There is a metallic ring of radius \(20\,\text{cm}\) placed in a uniform unsteady magnetic field \( B = (2t^2 + 2t + 3)\,\text{T} \) perpendicular to the plane of the ring. If the resistance of the ring is \(2\,\Omega\), find the induced current at \(t = 2\,\text{s}\). \([\pi = 22/7]\)

• A. \(0.63\,\text{A}\)
• B. \(0.063\,\text{A}\)
• C. \(0.3\,\text{A}\)
• D. \(6.3\,\text{A}\)

Hint:

When magnetic field varies with time but the loop is stationary: \[ \varepsilon = A\frac{dB}{dt} \] Always differentiate the magnetic field first and then substitute the time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Faraday's Law of Electromagnetic Induction, a changing magnetic field passing through an area induces an electromotive force (EMF) in a loop.
The induced current can then be calculated using Ohm's Law.
Step 2: Key Formula or Approach:
Induced EMF is given by $\varepsilon = \left| \frac{d\phi}{dt} \right| = A \left| \frac{dB}{dt} \right|$.
Induced current is $I = \frac{\varepsilon}{R}$.
Step 3: Detailed Explanation:
The area of the metallic ring is:
\[ A = \pi r^2 = \pi (0.2)^2 = 0.04\pi \text{ m}^2 \]
The magnetic field is given by $B = 2t^2 + 2t + 3$.
Differentiating with respect to time $t$:
\[ \frac{dB}{dt} = \frac{d}{dt}(2t^2 + 2t + 3) = 4t + 2 \]
At $t = 2 \text{ sec}$, the rate of change of magnetic field is:
\[ \left. \frac{dB}{dt} \right|_{t=2} = 4(2) + 2 = 10 \text{ T/s} \]
The induced EMF $\varepsilon$ at $t=2 \text{ sec}$ is:
\[ \varepsilon = A \frac{dB}{dt} = 0.04\pi \times 10 = 0.4\pi \text{ V} \]
The induced current $I$ is:
\[ I = \frac{\varepsilon}{R} = \frac{0.4\pi}{2} = 0.2\pi \text{ A} \]
Using $\pi = \frac{22}{7}$:
\[ I = 0.2 \times \frac{22}{7} = \frac{4.4}{7} \approx 0.628 \text{ A} \]
This rounds to $0.63 \text{ Amp}$.
Step 4: Final Answer:
The induced current at t = 2 sec is 0.63 Amp.