Question

A solenoid of length 40.0 cm is made of 9000 circular coils. It carries a steady current of 13.0 A . Near its center is placed a small, flat, circular metallic coil of 250 circular loops, each with a radius of 2.00 mm . This small coil is oriented so that it receives half of the maximum magnetic flux. A switch is opened in the solenoid circuit and its current drops to zero in 25.0 ms .

What was the initial flux through the small coil?

Determine the average induced emf in the small coil during the 25.0 ms .

If you look along the long axis of the solenoid so that the initial 13.0 A current is clockwise, determine the direction of the induced current in the small inner coil during the time the current drops to zero. (Clockwise or Counterclockwise)

During the 25.0 ms , what was the average current in the small coil, assuming it has a resistance of 0.200 Ω ?

Answer 1

Here ,

number of turns in coil , N = 250

For the initial flux in the small coil .

initial flux in the small coil = N * initial magnetic field * area * 0.5

initial flux in the small coil = 250 * u0 *(9000/.40) * 13 * pi *(0.002)^2 * 0.5

initial flux in the small coil = 5.77 *10^-4 T.m^2
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average induced emf = change in flux/time

average induced emf = (5.77 *10^-4 - 0).025

average induced emf = 0.0231 V

the average induced emf is 0.0231 V

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as the current is decreasing ,

the flux in the coil is decereasing ,

according to Len's law , the current in coil will flow in same direction as solenoid

hence , the current in coil is clockwise

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average current = average induced emf/R

average current = 0.0231/.2

average current = 0.115 A

the average current in the small coil is 0.115 A