Phys help show working please Consider two long, parallel metal rails, which are

Question

Phys help show working please

Consider two long, parallel metal rails, which are a distance D = 10 cm apart, as illustrated below. They are connected via a battery that supplies a potential difference Delta V= 100 V across them. The circuit is completed by a metal bar that is placed across the rails and which is free to slide left or right along the rails. In the problems below, ignore sliding friction and assume the resistance of the rails is negligible compared to the resistance (R = 10 Ohm) of the bar. The whole apparatus is placed in a region containing a uniform magnetic field of strength B = 1.0 T. In what direction should the uniform magnetic field be to generate the maximum acceleration of the bar to the right? Suppose that, at a particular instant, the current in the circuit is I = 8 A and the bar is Travelling with a velocity of v = 200 m/s to the right. What is the rate at which the magnetic field does work on the bar? Compare your result in (ii) to the power supplied by the battery. Comment on any difference you find. Is this result consistent with the conservation of energy? (For higher grades, this part must be answered in addition to ) Faraday's law predicts that the motion of the bar will induce an emf in the circuit. With the aid of a diagram, explain how this emf arises, making sure you identify the force that does the work on the charges. Derive an expression that describes how the induced emf depends on the velocity v of the bar. From Lenz's law, we expect that the induced emf will tend to oppose the force that creates it. Explain why, if the velocity of the bar is v = 200 m/s, then the current in the circuit is I = 8 A rather than the 10 A expected from I = Delta V/R. What is the maximum velocity that the bar can reach for the potential difference given above?

Explanation / Answer

i)the field should be coming out of the field of paper becuase then the induced current would be in clockwise direction and would add to the current due to battery. The force on bar is ILB, so when I is increased the force also increases.

ii)Force = ILB= 8*200*1 =1600N

power = force*velocity = -1600*200=-320000W

the power is negative becuase F and v are in opposite direction.

iii)Power supplied by battery=V*I = 100*8=800W

the difference is ude to the external force on the bar that needs to be applied and when the work by external force is taken into account then the conservation of energy is satisfied.

iv)flux in the area = D*x*B

rate of change of flux = DB*dx/dt=DB*v=0.10*1*200=20V

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