A solenoid of length 20.0 cm is made of 5000 circular coils. It carries a steady

Question

A solenoid of length 20.0 cm is made of 5000 circular coils. It carries a steady current of 10.0 A. Near it's center is placed a very flat and small coil with a resistance of 2.50 ohms made of 100 circular loops, each with a radius of 3.00 mm. This small coil is oriented so that its area receives the maximum magnetic flux. A switch is opened in the solenoid circuit and its current drops to zero in 15.0 ms. (a) What is the initial magnetic flux through the inner coil? (b) Determine the average induced emf in the small coil during the 15.0 ms. (c) If you look along the long axis of the solenoid so that the initial 10.0 A current is clockwise, determine the direction of the induced current in the small inner coil during the time that the current drops to zero. (d) What is the magnitude of the average induced current in the coil? A solenoid of length 20.0 cm is made of 5000 circular coils. It carries a steady current of 10.0 A. Near it's center is placed a very flat and small coil with a resistance of 2.50 ohms made of 100 circular loops, each with a radius of 3.00 mm. This small coil is oriented so that its area receives the maximum magnetic flux. A switch is opened in the solenoid circuit and its current drops to zero in 15.0 ms. (a) What is the initial magnetic flux through the inner coil? (b) Determine the average induced emf in the small coil during the 15.0 ms. (c) If you look along the long axis of the solenoid so that the initial 10.0 A current is clockwise, determine the direction of the induced current in the small inner coil during the time that the current drops to zero. (d) What is the magnitude of the average induced current in the coil?

Explanation / Answer

a.)B=u0*N*I=0.0628 T

flux=BAcos(0)=BA=0.0628*pi*(3x10^-3)^2= 1.776 x 10^-6 Wb

b.)E=N(d(flux))/dt=100*1.776x10^-6/15x10^-3= 0.01184 V

c.)the flux through the coil decreases when the switch is opened, so direction of induced current is clockwise.

d.)E=IR

I=0.01184/2.5= 4.736 x 10^-3 A

So B=u0*N*I= 5.95 x 10^-7 T

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