b) The emf around the loop causes a current to flow. How large is that current?

Question

b) The emf around the loop causes a current to flow. How large is that current? (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= 3.60 N along a set of rails separated by d= 0.62 m. A uniform magnetic field B= 0.750 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 3.52 m/s. a) Calculate the magnitude of the induced emf around the loop in the figure, due to Faraday's Law and the changing flux. b) The emf around the loop causes a current to flow. How large is that current? (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)?

Explanation / Answer

a)

Induced emf around the loop is

E=-B(dA/dt)=-B[(d/dt)(dvt)]

E=-Bdv =-0.75*0.62*3.52

E=-1.6368 volts =-1.64 volts

b)

The Force on the rod is given by

F=BId

Current in the loop

I=F/Bd =3.6/0.75*0.62

I=7.74 A

c)

Resistance

R=V/I =7.74/1.64

R=4.73 ohms

d)

Rate of energy disispation

P=F*v =3.6*3.52

P=12.67 Watts

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