A conducting rod is pulled horizontally with constant force F= 4.80 N along a se

Question

A conducting rod is pulled horizontally with constant force F= 4.80 N along a set of rails separated by d= 0.500 m. A uniform magnetic field B= 0.500 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 3.70 m/s.

first picture: http://www.flickr.com/photos/67913919@N05/

Using Faraday's Law, calculate the induced emf around the loop in the figure that is caused by the changing flux. Assign clockwise to be the positive direction for emf.

The rate at which the external force does mechanical work must be equal to the rate at which energy is dissipated in the circuit. What is that rate of energy dissipation (power dissipated)?

When the magnetic field is uniform and normal to the plane of the loop, then the flux is of the product of the field and the area. In this problem it is the area that changes with time, not the field.

Explanation / Answer

The induced voltage = B*L*v and the induced current comes from F = I*L*B or I = F/(L*B)

so R = Emf/I = B*L*v/(F/(L*B)) = (L*B)^2*v/F = (0.440*0.600)^2*4.60/3.60 = 0.0891%u03A9

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