need A conducting rod is pulled horizontally with constant force - 3.60 N along a set of rails separated by 0.520 m. A u

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Need A Conducting Rod Is Pulled Horizontally With Constant Force 3 60 N Along A Set Of Rails Separated By 0 520 M A U 1 (25.29 KiB) Viewed 45 times

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)?
need A conducting rod is pulled horizontally with constant force - 3.60 N along a set of rails separated by 0.520 m. A uniform magnetic field B-0.600 Tis directed into the page. There .. is no tinction between the rod and the rails and the rod moves with constant velocity v 5.40 ms. х х х х х х х X X F d х х х х х х х X X х Usina Faraday's Law, calculate the induced om around the loop in the figure that is caused by the changing flux. Assign clockwise to be the positive direction for em Tries 0/12 The emt around the loop causes a current to Hlow. 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 rails is negligible compared to the resistance of the rod, so the resistance of the loop is constanta w Tries 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