Solution Set<br>
Week 8<br>
<p>
Discusssion Questions:
<p>
9.  The transmission line generates a time-varying B-field which threads the
loop. The changing magnetic flux through the loop generates an <i>emf</i> and
so current flows through the loop. This would cause a slight drop in the
power delivered by the transmission line and thus would be detectable.
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16.  First off, the electron is forced to orbit in the magnetic field because
of the F = q(v x B) force it feels. That is, for a given B-field, 
it will move in a circular orbit
of radius r = mv/(qB) with frequency Omega = qB/m. Now if the B-field is
gradually increased in strength, the magnetic flux through the orbit increases.
This generates a current which opposes the increase in the B-field.  Looking at
the figure, we see that the B-field points down. 
The current will thus generate a B-field in the
upward direction. The current thus circulates in the CCW-sense as viewed
from above to generate such a field.  That is, the induced E-field points in the
CCW-direction as viewed from above.  The force on the electron due to this
induced field is then F = q(electron) x E which is then in the CW-direction
because the charge of the electron is negative. Now looking at the figure, we
see that the electron is moving in the CW-sense as viewed from above. The 
increasing magnetic field thus accelerates the electron to higher energy!
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Problems:
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2.  Flux = B x Area x cos(angle field makes with the perpendicular of the loop)
= 0.1 T x 0.020 m^2 x cos (30 degrees) = 0.0017 Wb
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Note, we can also say that 
Flux = B x Area x sin(angle field makes the plane of the loop)
= 0.1 T x 0.020 m^2 x sin(60 degrees) = 0.0017 Wb
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7. We want the average time rate of change of the magnetic flux, so
<ul>
<li>Initial: flux = 0.40 T x 0.25 m^2 = 0.1 Wb
<li>Final: flux = 0.0
</ul>
The time rate of change is then
<ul>
<li>delta(flux)/delta(time) = [final-initial]/0.2 s= 0.1 Wb/0.2 s = 0.5 Wb/s
</ul>
The induced <i>emf</i> is then -0.5 Volts
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8.  Again, we want the change in the flux:
<ul>
<li>Initial: flux = B x 100 x 0.0004 m^2 = 0.04 m^2 B
<li>Final: flux = 0
</ul>
Induced <i>emf</i> is
<ul>
<li><i>emf</i> = 1.0 V = -[0-0.04 B]/0.020 s ===> B = 0.020/0.040 T = 0.5 T
</ul>
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12.  let us consider the change in three steps:
<ul>
<li>Time = 0 s to 0.3 s:
<ul>
<li>Initial flux = 0 Wb
<li>Final flux = 0.3 T x 10 x 0.25 m^2 = 0.75 Wb
<li>===><i>emf</i>=-(0.05 Wb-0 wB)/0.3 s = -2.5 Volts
</ul>
<li>Time = 0.3 s to 0.4 s:
<ul>
<li>Since B is constant ===> no change in the flux and <i>emf</i> = 0.
</ul>
<li>Time = 0.4 s to 0.5 s:
<ul>
<li>Initial flux = 0.3 T x 10 x 0.25 m^2 = 0.75 Wb
<li>Final flux = 0.
<li>===><i>emf</i>=-(0 Wb-0.75 Wb)/0.1 s = 7.5 Volts
</ul>
</ul>
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16.  Since <i>emf</i> = vBl ===> B = <i>emf</i>/(vl) = 0.45 V/(6 m/s x 0.2 m) =
0.38 T.
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21.  In order to calculate the charge on the capacitor, we want to calculate
the induced <i>emf</i>.
<ul>
<li><i>emf</i> = - N x Area x (time rate of change of the B-field) = - 220 x
0.031 m^2 x (-0.020 T/s) = 0.14 Volts. We used the minus sign for the change in
the B-field because the strength of the field was decreasing.
</ul>
Now the charge on a fully-charged capacitor is given by
<ul>
<li>Q=CV=30x10^(-6) F x 0.14 V = 4.1 x 10^(-6) C
</ul>
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23.  Let us compute the time rate of change of the magnetic flux:
<ul>
<li>Initial flux = 0 Wb
<li>Final flux = 0.5 T x 0.0016 m^2 x cos(30 degrees) = 0.00069 Wb
<li>===> <i>emf</i>=-0.00069 Wb/0.200 s = 0.0035 Volts
</ul>
The current will flow in the CCW-direction as viewed from above so as to
oppose the increase in the downward B-field strength.
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25.  The induced <i>emf</i> is vBl ===> I = vBl/R ===> P = IV = vBl/R x vBl =
(vBl)^2/R
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33.  For each length of wire, the induced <i>emf</i> is vBl x sine(angle) = 22
m/s x 0.35 T x 0.1 m x  sine(angle) = 0.77 sine(angle) Volts. When the angle is 
90 degrees, the induced <i>emf</i> is 0.77 Volts. Now since there are 2 wires 
per turn and there are 25 turns, the total <i>emf</i> is 39 Volts.
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39.  N x Flux = LI ===> Flux = LI/N = 0.200 H x 2.0 A / 500 = 0.0004 Wb
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45.  Back <i>emf</i> = -L delta(I)/delta(time) ===> -L = 10 Volts/2.0 A/s = 5 H
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