Question:

A rod of length 60 cm rotates with a uniform angular velocity \(20 \, \text{rad} \, \text{s}^{-1}\) about its perpendicular bisector, in a uniform magnetic field \(0.5 \, \text{T}\). The direction of the magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is _____ V.

Updated On: Nov 24, 2024
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Solution and Explanation

The given data:
Length of rod, \(\ell = 60 \, \text{cm} = 0.6 \, \text{m}\), Angular velocity, \(\omega = 20 \, \text{rad/s}\), Magnetic field, \(B = 0.5 \, \text{T}\). 

The potential difference \(V_O - V_A\) between the ends of a rod rotating in a magnetic field is given by: 
\[ V_O - V_A = \frac{B \omega \ell^2}{2}. \] 
Substitute the values: 
\[ V_O - V_A = \frac{0.5 \times 20 \times (0.6)^2}{2} = \frac{0.5 \times 20 \times 0.36}{2}. \] 
\[ V_O - V_A = \frac{3.6}{2} = 1.8 \, \text{V}. \] 
However, since the magnetic field is parallel to the axis of rotation, no emf is induced across the rod. Therefore, 
\[ V_A = V_B \implies V_A - V_B = 0. \] 
Answer: \(0 \, \text{V}\)

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