JEE Physics 2024 Question with Solution

Given: Side of square loop = $10 \, \text{cm} = 0.1 \, \text{m}$, magnetic field = $0.20 \, \text{T}$, time interval = $1 \, \text{s}$. The loop is in the east-west vertical plane, so its area vector is along the north-south direction.

Find: The value of $x$ if induced emf is $\sqrt{x} \times 10^{-3} \, \text{V}$.

Area of the square loop:

$$A = (0.1)^2 = 0.01 \, \text{m}^2$$

The magnetic field is in the northeast direction, so the component perpendicular to the loop is:

$$B_\perp = B \cos 45^\circ = 0.20 \times \frac{1}{\sqrt{2}} = 0.1414 \, \text{T}$$

Magnetic flux through the loop initially is:

$$\phi = B_\perp A = 0.1414 \times 0.01 = 1.414 \times 10^{-3} \, \text{Wb}$$

Since the field decreases to zero uniformly in $1 \, \text{s}$, the magnitude of induced emf is:

$$\varepsilon = \frac{\Delta \phi}{\Delta t} = \frac{1.414 \times 10^{-3}}{1} = 1.414 \times 10^{-3} \, \text{V}$$

Now,

$$\varepsilon = \sqrt{x} \times 10^{-3} \, \text{V}$$

so,

$$\sqrt{x} = 1.414 = \sqrt{2}$$

Hence,

$$x = 2$$

Therefore, the value of $x$ is $2$.

Given: Side of square = $0.1 \, \text{m}$, field magnitude = $0.20 \, \text{T}$, time = $1 \, \text{s}$.

Find: The numerical value of $x$.

Take east as $\hat{i}$ and north as $\hat{j}$. Since the loop is in the east-west plane, its area vector is along north-south. Using the solution approach,

$$\vec{A} = 0.01 \, \hat{j} \, \text{m}^2$$

The magnetic field in northeast direction has equal east and north components:

$$\vec{B} = \frac{0.20}{\sqrt{2}} \, \hat{i} + \frac{0.20}{\sqrt{2}} \, \hat{j}$$

Flux is the dot product:

$$\Phi = \vec{B} \cdot \vec{A}$$ $$\Phi = \left(\frac{0.20}{\sqrt{2}} \, \hat{i} + \frac{0.20}{\sqrt{2}} \, \hat{j}\right) \cdot \left(0.01 \, \hat{j}\right)$$ $$\Phi = \frac{0.20}{\sqrt{2}} \times 0.01 = 1.414 \times 10^{-3} \, \text{Wb}$$

Therefore, induced emf magnitude is:

$$|\varepsilon| = \frac{\Phi}{1} = 1.414 \times 10^{-3} \, \text{V} = \sqrt{2} \times 10^{-3} \, \text{V}$$

Comparing with $\sqrt{x} \times 10^{-3} \, \text{V}$, we get $x = 2$.

Therefore, the final answer is $2$.