Question

A thin conducting square loop of side \(L\) is placed in the first quadrant of the \(xy\)-plane with one of the vertices at the origin. If a changing magnetic field \(\vec{B}(t) = B_0(52yt\,\hat{x} + xzt\,\hat{y} + 3y^2t^2\,\hat{z})\) is applied, where \(B_0\) is a constant, then the magnitude of the induced electromotive force in the loop is:

• A. \(4B_0L^4\)
• B. \(3B_0L^4\)
• C. \(2B_0L^4\)
• D. \(B_0L^4\)

Hint:

Always select the magnetic field component perpendicular to the surface for calculating flux through a loop.

Answer 1

The Correct Option is D

Step 1: Use Faraday's law of induction.
The induced emf is given by \[ \mathcal{E} = -\frac{d}{dt} \int \vec{B} \cdot d\vec{A} \]

Step 2: Consider only the \(z\)-component.
Since the loop lies in the \(xy\)-plane, the relevant magnetic field component is \(B_z = 3B_0 y^2 t^2\).

Step 3: Integrate over the loop area.
\[ \Phi = \int_0^L \int_0^L 3B_0 y^2 t^2 \, dx \, dy = 3B_0 L t^2 \int_0^L y^2 dy = B_0 L^4 t^2 \]

Step 4: Differentiate with respect to time.
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -2B_0 L^4 t \] At \(t = 1\), magnitude = \(3B_0 L^4.\)

Step 5: Conclusion.
Hence, the magnitude of induced emf is \(3B_0L^4.\)