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 0.36-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 8.4-s period after the rod has passed point A. V

Answer 1

Answer:

The emf = 1.53V

Explanation:

The end is given by Faraday'slaw as:

Emf = B dA/dt

Where B = 0.36T

dA/dt = rate of change of triangle's area.

Let's assume, it is an isosceles triangle with angle,

19-degree angle at the peak,

and the copper rod opposite that.

During the 8.4 second period after the rod passes point the intersection of the two rails,

the area of the triangle goes from nothing to 

[(0.6m/s)(8.4s)]^2 tan (19°/2)

Therefore, the average value of dA/dt = [0.6×8.4]^2 tan9.5°

dA/dt ave.= 25.40 × 0.1673

dA/dt = 4.25m^2/s

Emf = 0.36T × 4.25m^2/s

Emf = 1.53v