A conducting rod is pulled horizontally with constant force F= 3.60 N along a set of rails separated by d= 0.520 m. A un

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A conducting rod is pulled horizontally with constant force F= 3.60 N along a set of rails separated by d= 0.520 m. A uniform magnetic field B= 0.600 T is directed into the page. There
Is no rictio y, etwe gn the jod and the rally, and the rod moves with constant velocity v= 5.40 m/s.
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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.
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The emf around the loop causes a current to flow. How large is that current? (Again, use a positive value for clockwise direction.)
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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.)
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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)?
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A Conducting Rod Is Pulled Horizontally With Constant Force F 3 60 N Along A Set Of Rails Separated By D 0 520 M A Un 1 (26.15 KiB) Viewed 38 times

Cod A conducting rod is pulled horizontally with constant force Fs 3.60 N along a set of rails separated by 0.520 m. Aundommagnetic held 0.600 Tis directed into the page. There • is no friction between the rod and he ralls, and the rod moves with constant velocity V 5.40 m X х х х х х X х х F d X х х X х х х х х х х х Using Faraday's law, calculate the induced amt around the loop in the figure that is caused by the changing flux Assign clockwise to be the positive Grection for en Tres 0/12 The emt around the loop causes a current to flow. How large is that current? (Again, use a positive value for clockwise direction) Tries 0/12 from your previous results, what must be the electrical resistance of the loop? (The resistance of the railio a negligible compared to the resistance of the rod, so the resistance of the loop is Tres 0/12 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 so Tries 6/12 Post Discussion Send Feedback