A magnetic field of magnitude 9.8 Tesla passes through a square loop with side .3 meters, with the field at an angle of 9.4 degrees with a perpendicular to the plane of the loop. The loop is turned to make an angle of 4.6 degrees; the turn requires .008 seconds.
If the voltage is connected to a resistance of 8.6 `Ohms, then how much average power is required? Ignore the effects of self-induction in the loop, but explain what sort of effect self-induction might have on your result.
The area is .09 m ^ 2, so at 9.4 degrees the flux will be ( .09 m ^ 2)( 9.8 T)(cos( 9.4 deg)) = .87 T m ^ 2.
At 4.6 deg, a similar calculation yields flux .879 T m ^ 2.
The change in flux is thus .009 T m ^ 2.
Since this change occurs in .008 sec, the average rate of flux change is ( .009 T m ^ 2) / ( .008 sec) = 1.125 T m ^ 2 / sec 1.125 Volts.
Through 8.6 `Ohms this voltage would result in a current of .1308 amps, and a power of ( .1308 amps)( 1.125 Volts) = .1471 watts.
Self-induction comes about because the current in the loop is changing. A current in the loop results in an additional magnetic field in the vicinity of the loop, including inside the loop. As the current changes we therefore have a change in the magnetic field inside the loop, which results in a change in the voltage around the loop.
The flux at angle `theta from perpendicular will be B * A cos(`theta). The flux change between angles `theta1 and `theta2 will therefore be 4.6 `phi = B * A * ( cos(`theta2) - cos(`theta1) ).
If the flux change takes place in time interval `dt, then the voltage produced will be
V = 4.6 `phi / `dt = B * A ( cos(`theta2) - cos(`theta1) ) / `dt.
The resulting current will be I = V / R, and the resulting power will be P = I V = V^2 / R. The above expression for V may be substituted to obtain
The area is .09 m ^ 2, so at 9.4 degrees the flux will be ( .09 m ^ 2)( 9.8 T)(cos( 9.4 deg)) = .87 T m ^ 2.
At 4.6 deg, a similar calculation yields flux .879 T m ^ 2.
The change in flux is thus .009 T m ^ 2.
Since this change occurs in .008 sec, the average rate of flux change is ( .009 T m ^ 2) / ( .008 sec) = 1.125 T m ^ 2 / sec 1.125 Volts.
Through 8.6 `Ohms this will result in a current of .1308 amps, and a power of ( .1308 amps)( 1.125 Volts) = .1471 watts.
The flux at angle `theta from perpendicular will be B * A cos(`theta). The flux change between angles `theta1 and `theta2 will therefore be 4.6 `phi = B * A * ( cos(`theta2) - cos(`theta1) ).
If the flux change takes place in time interval `dt, then the voltage produced will be
V = 4.6 `phi / `dt = B * A ( cos(`theta2) - cos(`theta1) ) / `dt.
The resulting current will be I = V / R, and the resulting power will be P = I V = V^2 / R. The above expression for V may be substituted to obtain