Question

A conducting circular loop of area 1.0 m\(^2\) is placed perpendicular to a magnetic field which varies as B = sin(100t) Tesla. If the resistance of the loop is 100 \(\Omega\), then the average thermal energy dissipated in the loop in one period is_________ J.

• A. \(2\pi\)
• B. \(\pi\)
• C. \(\pi^2\)
• D. \(\pi/2\)

Hint:

For sinusoidal signals, the average value of \(\sin^2(\omega t)\) or \(\cos^2(\omega t)\) over a full period is always \(1/2\). You can quickly find the average power \( \langle P \rangle = \frac{\mathcal{E}_{rms}^2}{R} = \frac{(\mathcal{E}_0/\sqrt{2})^2}{R} = \frac{\mathcal{E}_0^2}{2R} \). Then, the total energy is simply \( E = \langle P \rangle \times T \). This avoids integration. Here, \( \langle P \rangle = \frac{100^2}{2 \cdot 100} = 50 \) W. \( E = 50 \times (\pi/50) = \pi \) J.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We have a changing magnetic field passing through a conducting loop, which will induce an EMF (and current) according to Faraday's Law. This current will dissipate energy as heat in the resistor. We need to find the total energy dissipated over one full time period of the magnetic field's oscillation.
Step 2: Key Formula or Approach:
1. Magnetic Flux: \( \Phi(t) = B(t) \cdot A \) (since the loop is perpendicular, \(\cos\theta = 1\)).
2. Induced EMF: \( \mathcal{E}(t) = - \frac{d\Phi}{dt} \).
3. Induced Current: \( I(t) = \frac{\mathcal{E}(t)}{R} \).
4. Instantaneous Power: \( P(t) = I(t)^2 R \).
5. Total Energy in one period: \( E = \int_0^T P(t) dt \), where T is the time period.
Step 3: Detailed Explanation:
Assuming the magnetic field is \(B(t) = \sin(100t)\) to match the given answer.
The angular frequency is \(\omega = 100\) rad/s.
The time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{100} = \frac{\pi}{50}\) s.
1. Calculate Flux and EMF:
Area \(A = 1.0\) m\(^2\).
\[ \Phi(t) = B(t) \cdot A = \sin(100t) \cdot 1 = \sin(100t) \, \text{Wb} \] \[ \mathcal{E}(t) = - \frac{d\Phi}{dt} = - \frac{d}{dt}(\sin(100t)) = -100 \cos(100t) \, \text{V} \] 2. Calculate Current and Power:
Resistance \(R = 100 \, \Omega\).
\[ I(t) = \frac{\mathcal{E}(t)}{R} = \frac{-100 \cos(100t)}{100} = -\cos(100t) \, \text{A} \] \[ P(t) = I(t)^2 R = (-\cos(100t))^2 \cdot 100 = 100 \cos^2(100t) \, \text{W} \] 3. Calculate Total Energy:
\[ E = \int_0^T P(t) dt = \int_0^{\pi/50} 100 \cos^2(100t) dt \] Using the identity \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\):
\[ E = 100 \int_0^{\pi/50} \frac{1 + \cos(200t)}{2} dt = 50 \left[ t + \frac{\sin(200t)}{200} \right]_0^{\pi/50} \] \[ E = 50 \left[ \left(\frac{\pi}{50} + \frac{\sin(200 \cdot \pi/50)}{200}\right) - (0 + 0) \right] \] \[ E = 50 \left[ \frac{\pi}{50} + \frac{\sin(4\pi)}{200} \right] \] Since \(\sin(4\pi) = 0\):
\[ E = 50 \left( \frac{\pi}{50} \right) = \pi \, \text{J} \] Step 4: Final Answer:
The average thermal energy dissipated in one period is \(\pi\) J. This corresponds to option (B).