A uniformly conducting semicircular wire rotates at constant angular speed ω in a uniform magnetic field B perpendicular to its plane. The rotation axis is parallel to B. When the rotation axis passes through end P, the potential difference between P and Q is 18 mV. If the axis is shifted to point R on the semicircular arc, with ∠POR = 60° as shown, find the new potential difference between P and Q (in mV).
Physics · Mixed Chapter
JEE Mains

Question

A uniformly conducting semicircular wire rotates at constant angular speed ω in a uniform magnetic field B perpendicular to its plane. The rotation axis is parallel to B. When the rotation axis passes through end P, the potential difference between P and Q is 18 mV. If the axis is shifted to point R on the semicircular arc, with ∠POR = 60° as shown, find the new potential difference between P and Q (in mV).

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Use the motional-potential relation: A wire rotating at angular speed ω in a uniform magnetic field B (parallel to its rotation axis) has a motional field. Between any two points on the wire: |ΔV| = (Bω/2) × |d₂² − d₁²|, where d₁ and d₂ are the respective distances of those points from the axis. The result does not depend on the shape of the wire joining them. Step 2 — When the rotation axis passes through P: Let a be the radius of the semicircle. Because PQ is its diameter, PQ = 2a. D…