Question

A horizontal telegraph wire of length $30~\text{m}$ fell from a height of $20~\text{m}$. If resistance is $40~\Omega$ and horizontal component of Earth’s magnetic field is $2 \times 10^{-5}~\text{T}$, the induced current is

• A. $0.3~\text{mA}$
• B. $3~\text{mA}$
• C. $3~\text{A}$
• D. $0.03~\text{A}$

Hint:

Use $v = \sqrt{2gh}$ and $emf = Blv$ to get induced current $I = emf/R$.

Answer 1

The Correct Option is A

$v = \sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 20} = \sqrt{400} = 20~\text{m/s}$
Induced emf $= Blv = 2 \times 10^{-5} \cdot 30 \cdot 20 = 0.012~\text{V}$
$I = \dfrac{E}{R} = \dfrac{0.012}{40} = 3 \times 10^{-4}~\text{A} = 0.3~\text{mA}$