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

To find the average thermal energy dissipated in the loop in one period, we need to follow these steps using Faraday's Law of Electromagnetic Induction and Ohm's Law.

1. Magnetic Flux: The magnetic flux \(\Phi_B\) through the loop is given by the product of the magnetic field \(B\) and the area \(A\) of the loop. Since the loop is placed perpendicular to the field, the flux is: \(\Phi_B = B \times A\). Given \(A = 1.0 \, \text{m}^2\), and \(B = \sin(100t)\) T, we have: \(\Phi_B = \sin(100t) \times 1 = \sin(100t)\) Weber.
2. Induced EMF: The electromotive force (EMF) \(\epsilon\) induced in the loop is given by Faraday's Law: \(\epsilon = -\frac{d\Phi_B}{dt}\). Differentiate \(\sin(100t)\) with respect to \(t\): \(\epsilon = -\frac{d}{dt}(\sin(100t)) = -100\cos(100t)\) V.
3. Current in the Loop: Using Ohm's Law, the current \(i\) in the loop is given by: \(i = \frac{\epsilon}{R}\), where \(R = 100 \, \Omega\). Thus: \(i = \frac{-100\cos(100t)}{100} = -\cos(100t)\) A.
4. Power Dissipation: The power \(P\) dissipated in the loop (as thermal energy) is given by: \(P = i^2 \times R\). Substitute \(i\): \(P = (-\cos(100t))^2 \times 100 = 100 \cos^2(100t)\) W.
5. Average Power Over One Period: To find the average power over one period, integrate \(P\) over the period \(T\) and divide by \(T\). The period \(T\) for \(\sin(100t)\) or \(\cos(100t)\) is: \(T = \frac{2\pi}{100}\). \(\text{Average Power } \langle P \rangle = \frac{1}{T} \int_0^T 100 \cos^2(100t) \, dt\). Using the trigonometric identity \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\): \(\int_0^T \cos^2(100t) \, dt = \int_0^T \frac{1 + \cos(200t)}{2} \, dt\). \(\langle P \rangle = \frac{100}{T} \int_0^T \frac{1 + \cos(200t)}{2} \, dt = \frac{100}{T} \left[\frac{1}{2} \int_0^T dt + \frac{1}{2} \int_0^T \cos(200t) \, dt\right]\). Evaluate the integrals: \(\int_0^T \cos(200t) \, dt = 0\) (since \(\cos\) of a complete period integrates to zero). \(\langle P \rangle = \frac{100}{T} \times \frac{T}{2} = 50\) W.
6. Energy Dissipated Over One Period: The average thermal energy dissipated, which equals the total energy dissipated over one period, is: \(E = \langle P \rangle \times T = 50 \times \frac{2\pi}{100} = \pi \) J.

Thus, the average thermal energy dissipated in the loop in one period is \(\pi\) J.