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Two long parallel conducting strips (or wires) of width/separation parameters carry equal and opposite currents I. For the standard parallel-plate current arrangement of separation b carrying surface current, the magnitude of the magnetic force per unit area between the two plates is:

1. mu0 * I² * l / (2*b)
2. mu0 * I² / (2*a²)
3. mu0 * I² / (2*b²)
4. mu0 * I² / (a*b)

Correct answer: mu0 * I² / (2*b²)

Solution

Each plate carries surface current density K = I/b. The field between two such sheets is B = mu0*K. The force per unit area on a current sheet is the surface current density times the average field it sits in, (1/2)*B: f = (1/2)*K*B = (1/2)*(I/b)*(mu0*I/b) = mu0*I²/(2*b²).

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