University

Physics 2

Physics 2

Fall 2024

Two loops of wire are stacked one above the other. The switch is open for a long time. Then, at time \(t=t_1\) the switch closes, and remains closed until all changes cease. Then, at time \(t=t_2\) the switch opens, and remains opened.

At time \(t=t_1^{+}\) (immediately after the switch is closed), and at time \(t=t_2^{+}\) (immediately after the switch opens again), determine the following:

- the direction of the induced current in the lower loop
- the direction of the induced electric field in the space around the lower loop
- the direction of the force on the lower loop

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

- counterclockwise
- counterclockwise
- downward (repulsive)

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

- clockwise
- clockwise
- 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)

A metal bar moves through the magnetic field shown.

Sketch the induced charges on bar.

A bar magnet is held near a conducting loop, which is connected to a light bulb. The magnet is able to move along the dotted line.

Give the current direction through the bulb (*up*,
*down*, or *none*) in the following cases.

- While the magnet is moving toward the loop.
- While the magnet is held stationary near the loop.
- While the magnet is moving away from the loop, toward the left.

- down
- none
- up

Reasons:

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

The magnetic field (pointing into the page) induces the electric field shown.

Complete the following statements by selecting the correct word.

- the electric field would push (
*a clockwise*/*a counterclockwise*/*no*) current if a conducting loop were present - the magnetic field is (
*increasing*/*decreasing*/*not changing*) in magnitude

- the electric field would push
*a counterclockwise*current if a conducting loop were present - the magnetic field is
*increasing*in magnitude

A wire loop is moved through the magnetic field, as shown.

At each lettered stage, determine

- the direction (or existence) of the induced current in the loop
- the direction (or existence) of the magnetic force on the loop

- no current, no force
- clockwise current, force to the left
- no current, no force
- counterclockwise current, force to the left

The figures below show one or more moving metal wires sliding through a magnetic field. For each, determine if the induced current flows clockwise, counterclockwise, or is zero.

- clockwise
- counterclockwise
- zero
- counterclockwise
- zero
- zero

The current \(I\) through a long solenoid with \(n\) turns per meter and radius \(R\) is a function of time given by \(I(t) = I_0 \sin \omega t\). Calculate the induced electric field as a function of distance \(r\) from the central axis of the solenoid.

Give your answer in terms of the variables given (and fundamental constants). Include radii inside and outside of the solenoid.

Answer:

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

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 \(\Phi = BA = \mu_0 n I \pi r^2\).

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

\(\displaystyle \frac{d\Phi}{dt} = \mu_0 n \pi r^2 \frac{dI}{dt} = \mu_0 n \pi r^2 I_0 \omega \cos \omega t\)

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

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:

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

Shown in the following figure is a long, straight wire and a single-turn rectangular loop, both of which lie in the plane of the page. The wire is parallel to the long sides of the loop and is \(0.50\) m away from the closer side.

At some moment, the emf induced in the loop is \(2.0\) V, with a counterclockwise current. At that moment, what is the time rate of change of the current in the wire? (Let the sign of \(dI/dt\) indicate if the current is increasing or decreasing.)

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: \(dA = w \; dy\).

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

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

The conducting rod shown in the accompanying figure moves along parallel metal rails that are \(25\) cm apart. The system is in a uniform magnetic field of strength \(0.75\) T, which is directed into the page. The resistances of the rod and the rails are negligible, but the section \(PQ\) has a resistance of \(0.25 \; \Omega\).

- What is the emf (including its sense) induced in the rod when it is moving to the right with a speed of \(5.0\) m/s?
- What force is required to keep the rod moving at this speed?
- What is the rate at which work is done by this force?
- What is the power dissipated in the resistor?

- \(0.94\) V, clockwise
- \(0.70\) N
- \(3.52\) J/s
- \(3.52\) 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.

A circular loop of wire of radius \(10\) cm is mounted on a vertical shaft and rotated at a frequency of \(5.0\) cycles per second in a region of uniform magnetic field of \(2\) Gauss perpendicular to the axis of rotation. The loop has a resistance of \(10\) Ω. At time \(t=0\) the loop has the orientation shown.

- Find an expression for the current through the loop as a function of time.
- What is the current (magnitude and direction) at time \(t=1.0\) s, when the loop is again in the orientation shown?

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

*Last modified: Tue April 16 2024, 11:46 AM.*