JEE Physics 2026 Question with Solution

Given: Length of wire $l = 20 \, \text{m}$, magnetic field $B = 0.5 \, \text{Gauss} = 0.5 \times 10^{-4} \, \text{T}$, acceleration due to gravity $g = 10 \, \text{m/s}^2$, and falling distance $h = 200 \, \text{m}$.

Find: The induced EMF across the wire in $\text{mV}$.

A conductor moving through a magnetic field experiences motional EMF given by

$$\varepsilon = Blv$$

where $v$ is the instantaneous speed perpendicular to the field and the length of the wire.

After falling through height $h$, the speed is

$$v = \sqrt{2gh}$$

Substituting the given values,

$$v = \sqrt{2 \times 10 \times 200} = \sqrt{4000} = 20\sqrt{10} \, \text{m/s}$$

Now use the motional EMF formula:

$$\varepsilon = (0.5 \times 10^{-4}) \times 20 \times (20\sqrt{10})$$ $$\varepsilon = 10^{-4} \times 200\sqrt{10} = 2 \times 10^{-2}\sqrt{10} \, \text{V}$$

Convert volts to millivolts using $1 \, \text{V} = 1000 \, \text{mV}$:

$$\varepsilon = 0.02\sqrt{10} \times 1000 = 20\sqrt{10} \, \text{mV}$$

Therefore, the induced EMF is $20\sqrt{10} \, \text{mV}$. The correct option is D.