JEE Physics 2026 Question with Solution

Given: mass of pendulum bob is $10 \, \text{g}$, length of metallic wire is $L = 10 \, \text{cm} = 0.1 \, \text{m}$, magnetic field is $B = 2 \, \text{T}$, release angle is $60^\circ$, and $g = 10 \, \text{m/s}^2$.

Find: the maximum induced EMF between the point of suspension and the point of oscillation.

The induced EMF is zero at the extreme positions because angular velocity is zero there. It is maximum at the mean position, where angular velocity is maximum.

For a rotating rod, the induced EMF between pivot and tip is

$$\varepsilon = \frac{1}{2}BL^2\omega$$

So we first find the maximum angular velocity using conservation of energy:

$$mgL(1-\cos\theta)=\frac{1}{2}m(L\omega_{\max})^2$$

Substituting $\theta=60^\circ$,

$$10 \times 0.1 \times (1-0.5)=\frac{1}{2}\times (0.1)^2 \times \omega_{\max}^2$$ $$0.5 = 0.005\,\omega_{\max}^2$$ $$\omega_{\max}^2=100$$ $$\omega_{\max}=10 \, \text{rad/s}$$

Now,

$$\varepsilon_{\max}=\frac{1}{2}BL^2\omega_{\max}$$ $$\varepsilon_{\max}=\frac{1}{2}\times 2 \times (0.1)^2 \times 10$$ $$\varepsilon_{\max}=0.1 \, \text{V}$$

Converting into millivolt,

$$0.1 \, \text{V} = 100 \, \text{mV}$$

Therefore, the maximum induced EMF is $100 \, \text{mV}$.