Set 54 Problem number 13

Problem

What average voltage is produced in a square loop with side .9 meters, which is positioned within a uniform magnetic field of magnitude 5 Tesla between the poles of a magnet, when the loop is turned from an angle of 3.4 degrees to an angle of 4.5 degrees, in .004 seconds?

• If the voltage is connected to a resistance of 7.4 `Ohms, then how much average power is required?  Ignore the effects of self-induction in the loop.
• Explain what sort of effect self-induction might have on your result.

Solution

The area is .81 m ^ 2, so at 3.4 degrees the flux will be ( .81 m ^ 2)( 5 T)(cos( 3.4 deg)) = 4.042 T m ^ 2.

At 4.5 deg, a similar calculation yields flux 4.037 T m ^ 2.

The change in flux is thus -.005001 T m ^ 2.

Since this change occurs in .004 sec, the average rate of flux change is (-.005001 T m ^ 2) / ( .004 sec) = -1.252 T m ^ 2 / sec -1.252 Volts.

Through 7.4 `Ohms this voltage would result in a current of -.1692 amps, and a power of (-.1692 amps)(-1.252 Volts) = .2115 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.  

• Lenz' Law tells us that this voltage must oppose any change in the current.  
• Since the voltage changes with the rotation of the loop, the current must also change.
• The voltage from self-induction must therefore oppose the voltage created in response to the main field, which results in less voltage than that computed.
• The lower the resistance of the circuit the more current tends to flow, and the more 'opposing' voltage is therefore created by self-induction within the loop.
• For the small currents in the present problem, self-induction is a negligible effect.

Generalized Solution

The flux at angle `theta from perpendicular will be B * A cos(`theta). The flux change between angles `theta1 and `theta2 will therefore be 4.5 `phi = B * A * ( cos(`theta2) - cos(`theta1) ).

If the flux change takes place in time interval `dt, then the voltage produced will be

• V = 4.5 `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

• power = V^2 / R = [ B * A (cos(`theta2) - cos(`theta1) ) / `dt ] ^2 / R = B^2 * A^2 / ( R `dt^2 ) * (cos(`theta2) - cos(`theta1) )^2.