Question

A wire of length \( 1 \, \text{m} \) moving with velocity \( 8 \, \text{m/s} \) at right angles to a magnetic field of \( 2 \, \text{T} \). The magnitude of induced emf between the ends of the wire will be:

• A. \( 20 \, \text{V} \)
• B. \( 8 \, \text{V} \)
• C. \( 12 \, \text{V} \)
• D. \( 16 \, \text{V} \)

Hint:

The induced emf in a conductor moving perpendicular to a magnetic field depends on the product of the magnetic field strength, the velocity of the conductor, and its length. Ensure all units are consistent when performing calculations.

Answer 1

The Correct Option is D

The induced emf \( \mathcal{E} \) in a conductor moving through a magnetic field is given by: \[ \mathcal{E} = B \cdot v \cdot \ell, \] where: \( B = 2 \, \text{T} \) is the magnetic field strength, \( v = 8 \, \text{m/s} \) is the velocity of the conductor, \( \ell = 1 \, \text{m} \) is the length of the conductor. Step 1: Substitute the Given Values Substitute \( B = 2 \, \text{T} \), \( v = 8 \, \text{m/s} \), and \( \ell = 1 \, \text{m} \) into the formula: \[ \mathcal{E} = 2 \cdot 8 \cdot 1. \] Step 2: After Simplify \[ \mathcal{E} = 16 \, \text{V}. \] Final Answer: \[ \boxed{16 \, \text{V}} \]