A long solenoid, with its axis along the x axis, consists of 150 turns per meter

Question

A long solenoid, with its axis along the x axis, consists of 150 turns per meter of wire that carries a steady current of 15.0 A. A coil is formed by wrapping 30 turns of thin wire around a circular frame that has a radius of 6.00 cm. The coil is placed inside the solenoid and mounted on an axis that is a diameter of the coil and coincides with the y axis. The coil is then rotated with an angular speed of 3.00 pi rad/s. (The plane of the coil is in the yz plane at t = 0.) Determine the emf generated in the coil as a function of time. (Use the following as necessary: t in seconds.) E = .1506 middot sin (3 pi middot t) mV

Explanation / Answer

magnetic field inside the long solenoid is:

B = µ0NI = (1.256x10-6T-m/A)(150/m)(15A) = 2.826x10-3T laong x-axis

Tunrs in the coil = n = 30

radius of the coil = 6cm

area of the coil is A = 3.14(6cm)2 = 0.0113m2

The coil is initially placed in yz plane at t=0. The area normal of the coil is initially parallel to the field in the x-direction.

Suppose the coil rotaes by an angle . Then the angle between the area normal and B in this position of the coil is .

So flux through the coil is = nB.A = nBAcos

so d/dt = -nBAsin.d/dt = -nBAsin (where = angular speed = 3rad/s)

So emf induced is E = -(d/dt) = nBAsin = nBAsint (As = t)

or E = nBAsint

or E = (30)(2.826x10-3T)(0.0113m2)(3rad/s)sin(3t)

or E = 2.874x10-3sin(3t)V = 2.874sin(3t)mV

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification I will be happy to oblige....

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