10 (part 1 of 4) 10.0 points A long solenoid carries a current I ( t ) = I 0 whe

Question

10 (part 1 of 4) 10.0 points A long solenoid carries a current I ( t ) = I 0 where I 0 = 30 A and = 1 . 6 s 1 . Inside this solenoid, and coaxial with it, is a smaller coil. The permeability of free space is 1 . 25664 × 10 6 Wb / A. 15 . 4 cm 5 . 4 cm Outside solenoid has 573 turns Inside coil has 380 turns 4 m 1 m Current I ( t ) What is the magnetic flux which passes through one of the turns of the smaller coil at t = 2 s? Answer in units of T m 2 . 011 (part 2 of 4) 10.0 points What is the mutual inductance? Answer in units of H. 012 (part 3 of 4) 10.0 points What is the magnitude of the maximum emf induced in the coil? Answer in units of V. 013 (part 4 of 4) 10.0 points If the axis of the coil in the figure is rotated an angle 38 from the horizontal, what is the new mutual inductance of the coil in the solenoid? Answer in units of H.

Explanation / Answer

If a current i in a coil is changing at a rate di/dt, then a voltage across the coil will be
induced in the direction to oppose the change. Vcoil = -L di/dt.
Similarly, if the time varying magnetic flux of one coil links another coil, then it will
induce a voltage in the 2nd coil of V'coil = -M di/dt. Where, i is the current in the 1st
coil.

The magnetic field in the larger coil is: B = µoNI/

The flux, , through the N' coil is BA, where A is the area of the 2nd coil.

= (µoNI/) r² = (720/4)(1.25664x10^-6)[25(1 - e^-1.6*2)] (0.089²)

= 1.35x10^-4 T/m²

Then, the voltage induced in the 2nd coil by the current in the 1st coil is:
V12 = -N' d/dt = - (µoNN' A/ )di/dt

The mutual inductance is: M = µoNN' A/

M = (1.25664x10^-6)720*340(0.089²)/4

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