1. Motional emf formula

For a rod of length $l$ moving with speed $v$ in a magnetic field, the induced emf is

$e = (\vec v \times \vec B) \cdot \vec l$

In magnitude form,

$e = B\,l\,v\sin\theta$

where $\theta$ is the angle between $\vec v$ and $\vec B$ in the effective geometry.

2. Given data

• Length of wire: $l = 10\,\text{m}$
• Speed of falling: $v = 5\,\text{m s}^{-1}$
• Horizontal component of earth's magnetic field: $B_H = 0.3 \times 10^{-4}\,\text{T}$

The wire lies along North-East to South-West.

The horizontal component of earth's magnetic field points along North.

The wire is falling vertically downward, so $\vec v$ is vertical and hence perpendicular to $\vec B_H$ (which is horizontal). Therefore, $|\vec v \times \vec B_H| = vB_H$.

Now the direction of $\vec v \times \vec B_H$ is along East-West.

Since the rod is along NE-SW, the angle between the rod and East-West direction is $45^\circ$.

Hence,

$e = B_H\,l\,v\cos 45^\circ$

or equivalently,

$e = B_H\,l\,v\sin 45^\circ$

since the relevant component along the rod is reduced by a factor $\frac{1}{\sqrt 2}$.

3. Substitute values

$e = (0.3 \times 10^{-4})(10)(5)\left(\frac{1}{\sqrt 2}\right)$

First,

$10 \times 5 = 50$

So,

$e = 0.3 \times 10^{-4} \times 50 \times \frac{1}{\sqrt 2}$

$= 15 \times 10^{-4} \times \frac{1}{\sqrt 2}$

$= 1.5 \times 10^{-3} \times \frac{1}{1.414}$

$\approx 1.06 \times 10^{-3}\,\text{V}$

$\approx 1.1 \times 10^{-3}\,\text{V}$

4. Match with options

The induced emf is

$\boxed{1.1 \times 10^{-3}\,\text{V}}$

So the correct option is D.

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.