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: Apply Faraday's Law

The induced EMF is: $$\mathcal{E} = -\frac{d\Phi_B}{dt}$$

where $\Phi_B = \iint_S \vec{B} \cdot d\vec{A}$

Step 2: Identify the loop geometry

The square loop in the first quadrant with one vertex at origin and side $L$ lies in the $xy$-plane (where $z = 0$).

The four vertices are: $(0,0,0)$, $(L,0,0)$, $(L,L,0)$, $(0,L,0)$

Area vector: $d\vec{A} = dx,dy,\hat{z}$

Step 3: Calculate magnetic flux

Since $z = 0$ on the loop: $$\vec{B}(t) = \beta_0(5(0)yt,\hat{x} + (0)xt,\hat{y} + 3y^2t\hat{z}) = \beta_0(3y^2t)\hat{z}$$

$$\Phi_B = \int_0^L \int_0^L \beta_0(3y^2t) ,dx,dy$$

$$= \beta_0 \cdot 3t \int_0^L y^2 ,dy \int_0^L dx$$

$$= \beta_0 \cdot 3t \cdot \frac{L^3}{3} \cdot L$$

$$= \beta_0 t L^4$$

Step 4: Calculate induced EMF

$$\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(\beta_0 t L^4) = -\beta_0 L^4$$

Magnitude: $$|\mathcal{E}| = \beta_0 L^4$$

Answer: (D) $\beta_0 L^4$