JEE Physics 2024 Question with Solution

Given: Magnetic field changes from $5000 \, \text{T}$ to $3000 \, \text{T}$ in $2 \, \text{s}$. Induced emf is $22 \, \text{V}$. Diameter of the coil is $0.02 \, \text{m}$.

Find: Number of turns $N$ in the coil.

Using Faraday's law of electromagnetic induction,

$$\varepsilon = -N \frac{\Delta \Phi}{\Delta t}$$

The magnetic flux through the coil is

$$\Phi = B A$$

Radius of the coil is

$$r = \frac{d}{2} = \frac{0.02}{2} = 0.01 \, \text{m}$$

So, area of the coil is

$$A = \pi r^2 = \pi (0.01)^2 = 0.0001\pi \, \text{m}^2$$

Change in magnetic field is

$$\Delta B = B_{\text{final}} - B_{\text{initial}} = 3000 - 5000 = -2000 \, \text{T}$$

Hence, change in magnetic flux is

$$\Delta \Phi = A \cdot \Delta B = 0.0001\pi \cdot (-2000) = -0.2\pi \, \text{Wb}$$

Substituting in Faraday's law,

$$22 = -N \cdot \frac{-0.2\pi}{2}$$ $$22 = N \cdot \frac{0.2\pi}{2}$$ $$22 = N \cdot 0.1\pi$$ $$N = \frac{22}{0.1\pi}$$ $$N \approx \frac{22}{0.314} \approx 70$$

Therefore, the number of turns in the coil is $70$. The correct option is B.

Given: $\mathcal{E} = 22 \, \text{V}$, $B_i = 5000 \, \text{T}$, $B_f = 3000 \, \text{T}$, $t = 2 \, \text{s}$, diameter $d = 0.02 \, \text{m}$.

Find: Number of turns $N$.

First calculate the radius and area:

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

Magnitude of change in magnetic field is

$$\Delta B = 2000 \, \text{T}$$

So,

$$\Delta \phi = (\Delta B)A = (2000)(\pi)(0.01)^2 = 0.2\pi$$

Now use

$$\mathcal{E} = N \left( \frac{\Delta \phi}{t} \right)$$ $$22 = N \left( \frac{0.2\pi}{2} \right)$$ $$N = 70$$

This works because the coil remains perpendicular to the magnetic field, so flux change depends directly on $BA$. Therefore, the correct option is B.