Problem 3 0.650 A conducting rod is pulled horizontally with constant force F 3.

Question

Problem 3 0.650 A conducting rod is pulled horizontally with constant force F 3.60 N along a set of rails separated by d-0.48 m. A uniform magnetic field B T is directed into the page. There is nofriction between the rod and the rails, and the rod moves with constant velocity v= 5.34 ms. a) Calculate the magnitude of the induced emf around the loop in the figure, due to Faraday's Law and the changing flux Submit Answer Tries 0/6 b) The emf around the loop causes a current to flow. How large is that current? (Use a positive value for clockwise direction.) Submit Answer Tries 0/6 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 Submit Answer Tries 0/6 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)? Submit Answer Tries 0/6 Due Friday April 06 11:59 am (EDT) B9

Explanation / Answer

According to the given problem,

Given d= 0.48 m ,v= 5.34 m/s ,B= 0.65 ,F=3.6N

=> Rate of change of area = d*v = 2.5632 m2/s
=> Rate of change of magnetic flux = B* (d*v) = 1.666 Weber/s
(a)

Since Flux is increasing (into the page)
Lenz' Law It tries to oppose the change so, using right hand thumb rule current must be produced in such a way to that B must be (from the page) induced current and hence EMF is in anti clock wise direction
=> its +1.666 V


(b)

Now since Force * Velocity = Power

We have P = 5.34 * 3.6

But P also equals V * I

=> 5.34 * 3.6 = 1.666 * I

=> I = 11.54 Amps


(c)

Now R = V/I

=> R = 1.666/11.54

=> R = 0.1444 Ohms


(d)

Power is same again

=> P= 5.34 * 3.6

=> P = 19.224 Watts.

I hope you understood the problem, If yes rate me!! or else comment for a better solution.

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