A solenoid of length 40.0 cm is made of 9000 circular coils. It carries a steady

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 ?

Explanation / Answer

magnetic field due to a solenoid = B = uo*(N/L)*i

magnetic flux = n*B*A*coswt

maximum flux = n*B*A

magnetic flux = half of the maximum magnetic flux

magnetic flux = nBA/2

the initial flux through the small coil = n*0.5*uo*(N/L)*i*pi*r^2

initial flux = 250*0.5*4*pi*10^-7*(9000/0.4)*13*pi*0.002^2

initial flux = 5.77*10^-4 Tm^2

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emf = flux/time = 5.77*10^-4 /0.025 = 0.2308 v

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clockwise

current = i = e/R = 0.2308/0.2 = 1.154 A

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