A rod of length 60 cm rotates with a uniform angular velocity 20 rad s⁻¹ about its perpendicular bisector, in a uniform magnetic filed 0.5 T. The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is _________ V.

Solution

Physics sir · 3 steps

Split the rod at the pivot: both halves are mirror images, so the two ends end up at the same potential.

1. Two half-rods. The rod turns about its midpoint O, so treat it as two rods of length l/2 = 0.3 m, each rotating about the same end O. The magnetic force on charges in each half points along the rod, either away from O or towards it, depending on the sense of rotation.

   V_O - V_(end) = 1/2Bω(l/2)² = 1/2 × 0.5 × 20 × (0.3)² = 0.45 V

2. Compare the ends. Both halves have the same length, B and ω, and in both the force points the same way relative to O. So each end differs from O by the same 0.45 V, with the same sign.
3. So. V_P - V_Q = (V_P - V_O) + (V_O - V_Q) = -0.45 + 0.45 = 0. The potential difference between the two ends is 0 V.

The common slip: 1.8. Using 1/2Bω l² for the full 0.6 m rod, as if it were pivoted at one end, gives 1.8 V. The pivot here is at the midpoint.

The idea to keep. For a rod rotating in a field, measure the EMF from the pivot outward. If the pivot is at the middle, the two halves give equal EMFs from centre to end, and the ends sit at the same potential.

Answer from the NTA final answer key. Solution by PrepWiser.