1. Given data

• Length of rod: $l = 1\,\text{m}$
• Angular velocity: $\omega = 5\,\text{rad s}^{-1}$
• Horizontal component of earth’s magnetic field: $B_H = 0.2 \times 10^{-4}\,\text{T} = 2 \times 10^{-5}\,\text{T}$

The rod rotates about one end in a vertical plane.

2. Formula for motional emf in a rotating rod

For a rod of length $l$ rotating with angular speed $\omega$ about one end in a magnetic field component perpendicular to the plane of rotation, the emf between its ends is

$\mathcal{E} = \frac{1}{2} B \omega l^2$

Since the rod rotates in a vertical plane, the magnetic field component perpendicular to this plane is the horizontal component of earth’s field, i.e. $B_H$.

So,

$\mathcal{E} = \frac{1}{2} B_H \omega l^2$

3. Substitute the values

$\mathcal{E} = \frac{1}{2}(2 \times 10^{-5})(5)(1)^2$

$\mathcal{E} = \frac{1}{2} \times 10^{-4}$

$\mathcal{E} = 5 \times 10^{-5}\,\text{V}$

4. Convert into microvolts

Since $1\,\mu\text{V} = 10^{-6}\,\text{V}$,

$5 \times 10^{-5}\,\text{V} = 50 \times 10^{-6}\,\text{V} = 50\,\mu\text{V}$

5. Match with options

• A: $5\,\text{mV}$
• B: $50\,\mu\text{V}$
• C: $5\,\mu\text{V}$
• D: $50\,\text{mV}$

Thus the correct option is:

$\boxed{\text{B }\; 50\,\mu\text{V}}$