Given data

• Length of wire: $l = 5\,\text{m}$
• Horizontal component of earth’s magnetic field: $B = 0.60 \times 10^{-4}\,\text{Wb m}^{-2} = 0.60 \times 10^{-4}\,\text{T}$
• Velocity of falling wire: $v = 10\,\text{m s}^{-1}$

Formula for motional emf

When a conductor of length $l$ moves with velocity $v$ perpendicular to magnetic field $B$, the induced emf is $\mathcal{E} = B l v$

Here:

• the wire is horizontal from east to west,
• it is falling vertically,
• the horizontal component of earth’s field is north-south.

So the wire, velocity, and magnetic field are mutually perpendicular. Hence, $\mathcal{E} = B l v$

Substitute the values $\mathcal{E} = (0.60 \times 10^{-4})(5)(10)$

$\mathcal{E} = 0.60 \times 50 \times 10^{-4}$

$\mathcal{E} = 30 \times 10^{-4}$

$\mathcal{E} = 3 \times 10^{-3}\,\text{V}$

Required integer

Since the emf is asked in the form $\underline{\hspace{2cm}} \times 10^{-3}\,\text{V}$ the required number is $3$

Comparison with stored answer

Stored correct answer = $3$

Our derived answer also = $3$.

Therefore, the answer agrees.