A square loop consists of a single turn with a resistance of 6.00 . The loop has

Question

A square loop consists of a single turn with a resistance of 6.00 . The loop has an area of 600 cm2, and has a uniform magnetic field passing through it that is directed out of the page. The loop contains a 12-volt battery, connected as shown in the figure below 12 V (a) At the instant shown in the figure, there is no net current in the loop. At what rate is the magnitude of the magnetic field changing? (b) If the polarity of the battery was reversed, and the magnetic field was still changing at the rate you calculated above, what would the magnitude of the net current through the loop be?

Explanation / Answer

Given

resistance in the circuit R = 6 ohm

Area of the square loop is A = 600 cm^2

battery of emf V = 12 V

no of turns N =1

a)

given there is no current , so that there is no net potential.

and we know that the change in magnetic flux = induced emf, here which is equal to the emf of the battery , in the opposite direction so that the potential is zero

that is  

e = - N dPhi/dt and  

the area of the square loop is constant only the field is changing so

so from the definition of magnetic flux phi = B*A cos theta = B*A hee theta = 0

12 V = -1*dphi/dt

12 V = -1*d(B*A)/dt

12 V = -1*A*dB/dt

dB/dt = 12/(-1*A) = -12/(0.06) = -200 T/s

so the rate of change of magnetic field is 200 T/s

b) if the polarity of the batterychanges then the the emf will be doubled (both emf will be in the same direction)

e = emf of battery + induced emf = 12+12 V = 24 V

so current from Ohm's law V = I*R ==> I = V/R = 24/6 = 4 A

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