Question

A conducting wire of length 2500 m is kept in east-west direction, at a height of 10 m from the ground. If it falls freely on the ground then the current induced in the wire is (Resistance of wire $= 25\sqrt{2} \;\Omega$, acceleration due to gravity $g=10\text{ m/s}^2$, $B_H=2\times10^{-5}\text{ T}$)

• A. 0.2 A
• B. 0.02 A
• C. 0.01 A
• D. 2 A

Hint:

Logic Tip: The orientation of the wire is critical. A wire falling horizontally while oriented East-West cuts the North-South magnetic field lines of the Earth, generating max EMF. If the wire were oriented North-South while falling, it would be parallel to the magnetic field lines, cutting nothing, and the induced EMF would be zero!

Answer 1

The Correct Option is B

Concept:
As a conducting wire falls freely under gravity, it cuts the horizontal component of the Earth's magnetic field ($B_H$). This induces a motional electromotive force (EMF) across the ends of the wire. The induced EMF ($e$) is given by $e = B_H l v$, where $v$ is the instantaneous velocity. The induced current ($I$) is then found using Ohm's Law: $I = \frac{e}{R}$.
Step 1: Calculate the velocity of the wire just before it hits the ground.
The wire falls freely from a height $h = 10\text{ m}$ with initial velocity $u = 0$. Using the third equation of motion ($v^2 = u^2 + 2gh$): $$v = \sqrt{2gh}$$ $$v = \sqrt{2 \times 10 \times 10} = \sqrt{200} = 10\sqrt{2}\text{ m/s}$$
Step 2: Calculate the induced EMF.
The wire is oriented East-West and falls downwards (vertically), so its velocity vector is perpendicular to the horizontal magnetic field component pointing North-South. $$e = B_H \cdot l \cdot v$$ Substitute the given values ($B_H = 2 \times 10^{-5}\text{ T}$, $l = 2500\text{ m}$): $$e = (2 \times 10^{-5}) \times 2500 \times (10\sqrt{2})$$ $$e = 2 \times 2500 \times 10 \times \sqrt{2} \times 10^{-5}$$ $$e = 50000 \times \sqrt{2} \times 10^{-5}$$ $$e = 5 \times 10^4 \times \sqrt{2} \times 10^{-5} = 0.5\sqrt{2}\text{ V}$$
Step 3: Calculate the induced current.
Using Ohm's law with the resistance $R = 25\sqrt{2} \;\Omega$: $$I = \frac{e}{R}$$ $$I = \frac{0.5\sqrt{2{25\sqrt{2$$ The $\sqrt{2}$ terms cancel out: $$I = \frac{0.5}{25} = \frac{1}{50}$$ $$I = 0.02\text{ A}$$