Question

A metallic rod of length 1 m held along east-west direction is allowed to fall down freely. Given horizontal component of earth’s magnetic field B_H = 3 × 10^-5 T. The emf induced in the rod at an instant t = 2s after it is released is ( Take g = 10 ms^-2 )

• A. 3 × 10^-3 V
• B. 3 × 10^-4 V
• C. 6 × 10^-3 V
• D. 6 × 10^-4 V

Answer 1

The Correct Option is D

Given Information: 
Length of rod, \( l = 1\,m \)
Horizontal component of Earth's magnetic field, \( B_H = 3 \times 10^{-5}\,T \)
Acceleration due to gravity, \( g = 10\,ms^{-2} \)
Time after release, \( t = 2\,s \)

Step-by-Step Explanation:

Step 1: Find velocity (\(v\)) of the rod after time \(t\):

Rod is falling freely under gravity, thus initial velocity \( u = 0 \), so velocity after \( t \) seconds is:

\[ v = u + gt = 0 + 10 \times 2 = 20\,m/s \]

Step 2: Calculate induced emf (\(E\)):

The induced emf in a rod moving perpendicular to a magnetic field is given by:

\[ E = B_H \times l \times v \]

Substitute values into this formula:

\[ E = (3 \times 10^{-5}\,T) \times (1\,m) \times (20\,m/s) \]

Simplify:

\[ E = 6 \times 10^{-4}\,V \]

Final Conclusion:
The induced emf in the rod at \( t = 2\,s \) is \(6 \times 10^{-4}\,V\).