A square conducting loop of side L contains two identical lightbulbs, 1 and 2. T

Question

A square conducting loop of side L contains two identical lightbulbs, 1 and 2. There is a magnetic field directed into the page in the region inside the loop with magnitude as a function of time t given by B (t) = at + b , where a and b are positive constants. The lightbulbs each have constant resistance R0 . Express all answers in terms of the given quantities and fundamental constants.

a. Derive an expression for the magnitude of the emf generated in the loop.

b. i. Determine an expression for the current through bulb 2.

ii. Indicate on the diagram above the direction of the current through bulb 2.

c. Derive an expression for the power dissipated in bulb 1.

Another identical bulb 3 is now connected in parallel with bulb 2, but it is entirely outside the magnetic field. Does the brightness of bulb 1 stay the same as before or does it get brighter or does it get dimmer. Justify your answer.

Next, the portion of the circuit containing bulb 3 is removed and a wire is added to connect the midpoints of the top and bottom of the original loop. How does bulb 1 compare to what was in the first circuit? Choose: 1)Brighter, 2) Dimmer, 3) the same - justify your answer.

Explanation / Answer

A)

flux=BA

induced emf is directly propotional to rate of change of flux

induced emf=-dpi/dt

=-A.dB/dt

B (t) = at + b

dB/dt=a

A is area of square loop=L2

INDUCED EMF=aL2

B)

two bulbs are connected in series so total resistance=2R0

current through bulb 2=E/2R0  =aL2/2R0

C)

power dissipitated in bulb1=I2R

=(aL2/2R0)2*R0

when third resistance is connected in parllel to bulb2

resultant resistance decreases so current increases voltage drop is high bulb will be brighter

in this case bright ness and darkness depends resistance of wire

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