Question

A metal wire of length $2500\ \text{m}$ is kept in east-west direction, at a height of $10\ \text{m}$ from the ground. If it falls freely on the ground then the current induced in the wire is (Resistance of wire $= 25\sqrt{2}\ \Omega, g = 10\ \text{m/s}^2$ and Earth's horizontal component of magnetic field $B_H = 2 \times 10^{-5}\ \text{T}$)

• A. $0.2\ \text{A}$
• B. $0.02\ \text{A}$
• C. $0.01\ \text{A}$
• D. $2\ \text{A}$

Hint:

If the wire was oriented North-South instead of East-West and fell vertically, it would be moving parallel to the horizontal magnetic field lines. It wouldn't "cut" them, so the induced emf (and current) would be zero!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A wire falling vertically cuts through the horizontal component of the Earth's magnetic field, inducing an electromotive force (emf). We need to calculate the current induced in it just before it hits the ground.

Step 2: Key Formula or Approach:
1. The velocity of a freely falling body is given by kinematics: $v = \sqrt{2gh}$.
2. The motional emf induced is $e = Blv$.
3. The induced current is $I = \frac{e}{R}$.

Step 3: Detailed Explanation:
Given parameters:
Length $l = 2500\ \text{m}$
Height $h = 10\ \text{m}$
Resistance $R = 25\sqrt{2}\ \Omega$
Magnetic field $B_H = 2 \times 10^{-5}\ \text{T}$
First, find the maximum velocity $v$ just before it reaches the ground:
$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = \sqrt{200} = 10\sqrt{2}\ \text{m/s}$$
Now, calculate the induced emf $e$:
Since the wire is East-West and falls vertically, it perfectly cuts the South-North horizontal magnetic field lines. Thus, all three vectors (velocity, length, B-field) are mutually perpendicular.
$$e = B_H l v$$ $$e = (2 \times 10^{-5}) \times 2500 \times (10\sqrt{2})$$ $$e = (2 \times 2500 \times 10\sqrt{2}) \times 10^{-5}$$ $$e = 50000\sqrt{2} \times 10^{-5}$$ $$e = 0.5\sqrt{2}\ \text{V}$$
Finally, calculate the induced current $I$:
$$I = \frac{e}{R} = \frac{0.5\sqrt{2}}{25\sqrt{2}}$$ $$I = \frac{0.5}{25} = \frac{1}{50} = 0.02\ \text{A}$$

Step 4: Final Answer:
The current induced is $0.02\ \text{A}$, matching option (B).