Question

If the emf produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amperes·meter at a speed of 4.20 meters/second, what is the length of the wire in the magnetic field?

Answer 1

Emf = d (phi-B) / dt 
B dA/dt, where dA/dt is the area swept out by the wire per unit time. 
0.88 V = (0.075 N/(A m)) (L)(4.20 m/s), so 
L = (0.88 J/C) / [ (0.075 N s/C m)(4.2 m/s) ] = about 3 meters

Answer 2

Emf = -d(flux)/dt flux = B A, field times area normal to field B is constant but A is changing, as L sweeps along, dA/dt = L (4.20 m/s) emf = - B dA/dt