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Two long parallel straight wires lie on a frictionless table, and two more parallel wires rest across them at right angles, initially forming a square of side a. A uniform magnetic field B is perpendicular to the plane of the wires. All four wires begin moving outward with constant speed v. If r is the resistance per unit length of the wire, what current flows in the circuit?

1. Bv/r
2. Br/v
3. Bvr
4. Bv

Correct answer: Bv/r

Solution

Let the side be s. Each pair of opposite wires separates at relative speed 2v, but treating the standard result: emf = B*dA/dt. With area A = s² and ds/dt giving dA/dt, and the loop resistance = r*(perimeter) = r*4s, the s-dependence cancels and the current becomes independent of size, equal to Bv/r. This is the classic result for such an expanding square.

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