A conducting rod is pulled horizontally with constant force F= 4.10 N along a se

Question

A conducting rod is pulled horizontally  with constant force F= 4.10 N along a set of rails separated by d= 0.580  m. A uniform magnetic field B= 0.700 T is directed into the page. There  is no friction between the rod and the rails, and the rod moves with  constant velocity v= 3.90 m/s.

(a) 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.

(b) The emf around the loop causes a current  to flow. How large is that current? (Again, use a positive value for  clockwise direction.)
          

(c) From your previous results, what must be  the electrical resistance of the loop?  (The resistance of the rails is  negligible compared to the resistance of the rod, so the resistance of  the loop is constant.)

(d) 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)?

A conducting rod is pulled horizontally with constant force F= 4.10 N along a set of rails separated by d= 0.580 m. A uniform magnetic field B= 0.700 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 3.90 m/s. 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 emf around the loop causes a current to flow. How large is that current? (Again, use a positive value for clockwise direction.) From your previous results, what must be the electrical resistance of the loop? (The resistance of the rails is negligible compared to the resistance of the rod, so the resistance of the loop is constant.) 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)?

Explanation / Answer

a) emf = B*V*L

= 0.7*3.9*0.58

= 1.5834 Volts

b) let i is the current flowing

p = emf*i

F*v = emf*i

==> i = F*v/emf

       = 4.1*3.9/1.5834

       = 10.098 A
direction of induced current is clockwise

c) R = emf/i = 1.5834/10.098 = 0.1568 ohms

d)

power dissipated = i^2*R = 10.098^2*0.1568 = 15.99 Watts

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