A long solenoid carries a current of 30 A. Another coil (of larger diameter than

Question

A long solenoid carries a current of 30 A. Another coil (of larger diameter than the solenoid) is coaxial with the center of the solenoid, as in the figure. Find the magnetic field in the solenoid. The outside coil has 100 turns and the inside solenoid has 7950 turns. The permeability of free space is 4 pi x 10-7 T. m / A. Answer in units of T Find the magnetic flux through the cross- sectional area of the large circular coil due to the 30 A current in the solenoid. Answer in units of T m2 Calculate the mutual inductance M of the solenoid and circular coil.

Explanation / Answer

For solenoids: B = µ?NI/L But here you have 2 solendoids, and the inner solenoid will induce a current in the outer solenoid, thus changing the NET magnetic field. The inductance relationship is given by: I2 = I1(N1/N2) = (30) (4200/300) = 420 A So the 2nd solenoid generates: B2 = (4px10^7)(300)(420) / (2) = 7.92 x 10^12 T However the induced current in the outer solenoid would be flowing in the opposite direction from the inner solenoid, so B2 would be in the opposite direction as well. So B(net) = 7.92 x 10^12 - 2.26x10^12 = 5.66x10^12 T

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