Question

A copper rod is sliding on two conducting rails that make an angle of 19° with respect to each other, as in the drawing. The rod is moving to the right with a constant speed of 0.60 m/s. A 0.29-T uniform magnetic field is perpendicular to the plane of the paper. Determine the magnitude of the average emf induced in the triangle ABC during the 8.5-s period after the rod has passed point A.

Answer 1

By Faraday's Law, the emf is
B dA/dt,
where B = 0.2 T and dA/dt is the rate of change of the triangle's area.

I'm going to assume the triangle is isosceles, with the 19-degree angle at the peak,
and the copper rod opposite that.

During the 8.5 second period after the rod passes point the intersection of the two rails,

the area of the triangle goes from nothing to [(0.60 m/s)(8.5 s)]^2 tan(19 degrees/2),

so

the average value of dA/dt during that time is

(0.60 m/s)^2 (8.5s) tan (8.9 deg) = 0.2740 m^2/s

Hence

, emf = (0.29 T)(0.240 m^2/s) = (0.29V s/m^2) (0.2740 m^2/s) = 0.794Volts