Ch 13 Model Questions

At time \(t=t_1^{+}\) (immediately after the switch is closed):

1. counterclockwise
2. counterclockwise
3. downward (repulsive)

At time \(t=t_2^{+}\) (immediately after the switch is opened):

1. clockwise
2. clockwise
3. upward (attractive)

Reasoning

At time \(t=t_1^{+}\):

• a clockwise current rapidly begins to grow in the upper loop.
• a downward-pointing magnetic field begins to grow in the space around the lower loop
• by Lenz’s Law, a magnetic field pointing up is induced in the lower loop
• this requires a counterclockwise induced current in the lower loop
• the counterclockwise induced current is due to a counterclockwise induced electric field in the space around the lower loop
• opposite currents will repel (or equivalently, the upper loop’s N pole repels the lower loop’s N pole)

At time \(t=t_2^{+}\):

• the clockwise current rapidly begins to decrease in the upper loop.
• the downward-pointing magnetic field begins to decrease in the space around the lower loop
• by Lenz’s Law, a magnetic field pointing down is induced in the lower loop
• this requires a clockwise induced current in the lower loop
• the clockwise induced current is due to a clockwise induced electric field in the space around the lower loop
• like currents will attract (or equivalently, the upper loop’s N pole attracts the lower loop’s S pole)

1. down
2. none
3. up

Reasons:

1. Rightward \(B\) field, getting stronger, induces clockwise current in loop, which is downward through the bulb.
2. No induction is there is no relative motion.
3. Rightward \(B\) field, getting weaker, induces counterclockwise current in loop, which is upward through the bulb.

1. the electric field would push a counterclockwise current if a conducting loop were present
2. the magnetic field is increasing in magnitude

1. no current, no force
2. clockwise current, force to the left
3. no current, no force
4. counterclockwise current, force to the left

1. clockwise
2. counterclockwise
3. zero
4. counterclockwise
5. zero
6. zero

Answer:

\[ E(r) = \left\{ \begin{array}{rcl} -\tfrac{1}{2} \mu_0 n r I_0 \omega \cos \omega t & \mbox{for} & r \lt R \\ -\frac{1}{2r} \mu_0 n R^2 I_0 \omega \cos \omega t & \mbox{for} & r \gt R \\ \end{array}\right. \]

(The negative signs mean the electric field loops in a direction opposite the counter-clockwise direction.)

Faraday’s Law gives the induced electric field at radius \(r\):

\(\displaystyle \oint \vec{E} \cdot d\vec{\ell} = -\frac{d\Phi}{dt}\)

Pick a loop of radius \(r\) around the axis of the solenoid.

The magnetic field of the solenoid is \(B = \mu_0 n I\). Magnetic flux through the loop of radius \(r\) is \(\Phi = \vec{B} \cdot \vec{A} = BA = \mu_0 n I \pi r^2\).

We will need the time derivative of the flux, which is

\[ \frac{d\Phi}{dt} = \mu_0 n \pi r^2 \frac{dI}{dt} = \mu_0 n \pi r^2 I_0 \omega \cos \omega t \]

The integral around the loop is \[ \oint \vec{E} \cdot d\vec{\ell} = E(2 \pi r) \]

For \(r \lt R\), put these together and solve for \(E\) to give

\[ E = -\frac{1}{2} \mu_0 n r I_0 \omega \cos \omega t \]

For \(r \gt R\), the loop is outside the solenoid. Now the flux is \(\Phi = \mu_0 n I \pi R^2\) (the entire solenoid) but the loop integral is still \(E(2 \pi r)\).

So:

\[ E = - \frac{1}{2r} \mu_0 n R^2 I_0 \omega \cos \omega t \]

Answer: \(\displaystyle \frac{dI}{dt} = 4.81 \times 10^6\) A/s

(Positive implies \(I\) is increasing in time.)

Use Faraday’s Law in the form \(\displaystyle \mathcal{E} = - \frac{d\Phi}{dt}\)

An emf that drives a counterclockwise current implies that the integral \(\mathcal{E} = \oint \vec{E} \cdot d\vec{\ell}\) circulates in the counterclockwise direction.

Since the magnetic field in the rectangular loop is different at different locations in the loop, flux must be found by integrating

\(\displaystyle \Phi= \int_{\rm{area}} \vec{B} \cdot d\vec{A}\)

Since the loop integral circulates counterclockwise, this implies that \(d\vec{A}\) points out of the page. The magnetic field points into the loop, so the dot product \(\vec{B} \cdot d\vec{A} = -B \; dA\).

So \(\displaystyle \Phi = - \int_{\rm{area}} B \; dA\)

Establish a coordinate system that has \(+x\) and \(+y\) down from the straight wire. Let \(w= 3.0\) m, the width of the rectangle. Then the area element \(dA\) can be made into a several thin horizontal rectangles so we can integrate over a single variable: \(dA = w \; dy\).

\(\displaystyle \Phi = - \int_{\rm{area}} B \; dA = - \int_{0.5}^{1.0} B(y) w \; dy = - w \int_{0.5}^{1.0} B(y) \; dy\)

Note: If you’re familiar with two-dimensional integrals, you could write the area element as \(dA = dy\;dx\), then integrate twice:

\(\displaystyle \Phi = - \int_{\rm{area}} B \; dA = - \int_{x=0}^{w} \; \int_{y=0.5}^{1.0} B(y) \; dy \; dx = - \int_{x=0}^{w} dx \; \int_{0.5}^{1.0} B(y) \; dy = - w \int_{0.5}^{1.0} B(y) \; dy\)

The magnetic field is that of a long straight wire:

\(\displaystyle B_{\rm{wire}} = \frac{\mu_0 I}{2 \pi y}\)

\(\displaystyle \Phi= - w \frac{\mu_0 I}{2 \pi} \int_{0.5}^{1.0} \frac{dy}{y} = - \frac{\mu_0 I w}{2 \pi} \left. \ln y \right \vert_{0.5}^{1.0} = - \frac{\mu_0 I w}{2 \pi} \ln 2\)

Now returning to Faraday’s Law:

\(\displaystyle \mathcal{E} = - \frac{d\Phi}{dt} = \frac{d}{dt} \left ( \frac{\mu_0 I w}{2 \pi} \ln 2 \right ) = \frac{\mu_0 w \ln 2}{2 \pi} \frac{dI}{dt}\)

So, solving for \(\frac{dI}{dt}\):

\(\displaystyle \frac{dI}{dt} = \mathcal{E} \frac{2 \pi}{\mu_0 w \ln 2} = 4.81 \times 10^6\) A/s

1. \(0.94\) V, clockwise
2. \(0.70\) N
3. \(3.52\) J/s
4. \(3.52\) W

Let the loop have dimensions \(h = 0.25\) m and \(w = w_0 - vt\), where \(w_0\) is some unknown intial width, and \(v = 5.0\) m/s.

\(\displaystyle \mathcal{E} = - \frac{d\Phi}{dt} = - \frac{d}{dt} \left [ Bh(w_0-vt) \right ] = Bhv = 0.938\) V

With resistance \(R\), current will be \(I=\mathcal{E}/R = 3.75\) A.

Force is \(F=BIh= 0.703\) N.

Rate of work done (and power dissipated) is \(P=Fv= 3.516\) W.

1. \(\displaystyle I(t) = - \frac{B \omega \pi r^2}{R} \cos \omega t\) \(= -(1.974 \times 10^{-5}\) A) \(\cos (10 \pi t)\)
2. \(I(t=1) = 1.974 \times 10^{-5}\) A, clockwise

Establish a coordinate system that has \(+x\) to the right, \(+y\) upward and \(+z\) out of the page.

Then \(\vec{B} = B \hat{\imath}\).

With \(\omega= 5 \cdot 2 \pi\) rad/s, the area vector is \(\vec{A} = \pi r^2 (\hat{\imath} \sin \omega t + \hat{k} \cos \omega t)\)

(Note that at time \(t=0\), this points in the \(\hat{k}\) direction, as shown in the figure.)

The flux is \(\displaystyle \Phi= \vec{B} \cdot \vec{A} = B \pi r^2 \sin \omega t\)

And

\(\displaystyle \mathcal{E} = - \frac{d\Phi}{dt} = - B \pi r^2 \frac{d}{dt} \left ( \sin \omega t \right ) = - B \omega \pi r^2 \cos \omega t\)

With resistance \(R\), current will be \(I=\mathcal{E}/R\), so

\(\displaystyle I(t) = - \frac{B \omega \pi r^2}{R} \cos \omega t\) \(= -(1.974 \times 10^{-5}\) A) \(\cos (10 \pi t)\)

At time \(t=1.0\) s, \(\cos (10 \pi t) = 1\), so the current is negative. Since \(\vec{A}\) points out of the page at this moment, a positive circulation would be counterclockwise (by the right-hand rule).