Question

A conducting rod is pulled horizontally with constant force F= 3.90 N along a set of rails separated by d= 0.220 m. A uniform magnetic field B= 0.800 T is directed, into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 3.80 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. 6.69×10-1 v You are correct. Previous Tries The emf around the loop causes a current to flow. How large is that current? (Again, use a positive value for clockwise direction.) 1.07m Remember F=ma: since v is constant, a=0, so it must be true that the net force on the rod is zero. The pulling force is compensating the force on the rod due to the current through it and the magnetic field. Submit Answer Incompatible units. No conversion found between "m" and the required units. Tries 7/20 Previous Tries t must be the electrical resistance of the loop? (The resistance of the rails is negligible From your previous results compared to the resistancé of the rod, so the resistance of the loop is constant 0.01364 ohms Use Ohms Law Submit Answer Incorrect. Tries 3/20 Previous Tries e rate, at which the external force does mechanical work must be equal to the rate at which energy is dissipated in the circuit at is that rate of energy dissipation (power dissipate You can either calculate the mechanical power generated by the external force, or the electrical power dissipation in the circuit. Do it both ways to check your answer Submit Answer Incorrect. Tries 1/20 Previous Tries

Answer 1

induced emf = B L v

= (0.800) (0.220) (3.80)

= 0.669 Volt

field is into the page and flux is increasing.

hence induced current will be counterclockwise

ANs: - 0.669 V  

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F = IL x B

3.90 = (I x 0.220) (0.800)

I = - 22.2 A ....Ans

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R = V / I

= 0.669 / 22.2

R = 0.03 ohm .....Ans

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P = F v = 3.90 x 3.80 = 14.82 Watt .....Ans

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P = I^2 R =22.2^2 x 0.03 = 14.8 Watt ...Ans