Question

A circular loop of wire with radius R is centered at the origin of the xy-plane. The magnetic field at a point within the loop is, \(\overrightarrow B(\rho,\phi,z,t)=k\rho^3t^3\hat{z}\), where k is a positive constant of appropriate dimensions. Neglecting the effects of any current induced in the loop, the magnitude of the induced emf in the loop at time t is

• A. \(\frac{6\pi kt^2R^5}{5}\)
• B. \(\frac{5\pi kt^2R^5}{6}\)
• C. \(\frac{3\pi kt^2R^5}{2}\)
• D. \(\frac{\pi kt^2R^5}{2}\)

Answer 1

The Correct Option is A

To find the magnitude of the induced emf in the loop, we need to use Faraday's law of electromagnetic induction, which states that the induced emf \((\mathcal{E})\) in a closed loop is equal to the negative rate of change of magnetic flux through the loop:

\(\mathcal{E} = - \frac{d\Phi_B}{dt}\)

Here, \(\Phi_B\) is the magnetic flux through the loop, which is given by:

\(\Phi_B = \int \overrightarrow{B} \cdot d\overrightarrow{A}\)

Since the magnetic field \(\overrightarrow{B}\) is given as \(\overrightarrow{B}(\rho,\phi,z,t) = k\rho^3t^3\hat{z}\) and the area element is \(d\overrightarrow{A} = \hat{z} \, \rho \, d\rho \, d\phi\) for a circular loop in polar coordinates, the magnetic flux \(\Phi_B\) becomes:

\[\Phi_B = \int_0^{R} \int_0^{2\pi} (k\rho^3t^3) \rho \, d\rho \, d\phi\]

This simplifies to:

\[\Phi_B = k t^3 \int_0^{2\pi} d\phi \int_0^{R} \rho^4 \, d\rho\]

First, we integrate with respect to \(\rho\) from \(0\) to \(R\):

\[ \int_0^{R} \rho^4 \, d\rho = \frac{\rho^5}{5} \Bigg|_0^R = \frac{R^5}{5} \]

Now, integrate with respect to \(\phi\) from \(0\) to \(2\pi\):

\[\int_0^{2\pi} d\phi = 2\pi\]

Thus, the magnetic flux \(\Phi_B\) becomes:

\[ \Phi_B = k t^3 \cdot 2\pi \cdot \frac{R^5}{5} = \frac{2\pi k t^3 R^5}{5} \]

The induced emf \(\mathcal{E}\) is then the negative derivative of the magnetic flux with respect to time:

\[\mathcal{E} = - \frac{d}{dt}\left(\frac{2\pi k t^3 R^5}{5}\right)\]

Calculating the derivative:

\[ \frac{d}{dt}(t^3) = 3t^2 \]

Hence, the induced emf is:

\[ \mathcal{E} = -\left(\frac{2\pi k R^5}{5} \times 3t^2\right) = -\frac{6\pi k t^2 R^5}{5} \]

The magnitude of the induced emf is:

\(\frac{6\pi k t^2 R^5}{5}\)

Thus, the correct answer is \(\frac{6\pi kt^2R^5}{5}\).