AP Physics C E and M Unit 12 Magnetic Fields

[Maximum number: 10]

Figure for Question 12 Magnetic Fields and Electromagnetism question 1 — AP Physics C: Electricity & Magnetism

A solenoid is used to generate a magnetic field. The solenoid has an inner radius a, length $\ell$ , and N total turns of wire. A power supply, not shown, is connected to the solenoid and generates current I, as shown in the figure on the left above. The x-axis runs along the axis of the solenoid. Point P is in the middle of the solenoid at the origin of the x y z-coordinate system, as shown in the cutaway view on the right above. Assume $\ell \gg a$ .

(a)

Select the correct direction of the magnetic field at point P. +x-direction +y-direction +z-direction - x-direction -y-direction -z-direction
Justify your selection.

[ 2 ]

(b)

On the cutaway view below, clearly draw an Amperian loop that can be used to determine the magnetic field at point P at the center of the solenoid.

Cutaway View

Use Ampere's law to derive an expression for the magnetic field strength at point P. Express your answer in terms of $I, \ell, N, a$ , and physical constants, as appropriate.

Some physics students conduct an experiment to determine the resistance $R_{S}$ of a solenoid with radius $a=0.015 \mathrm{~m}$ , total turns N=100, and total length $\ell=0.40 \mathrm{~m}$ . The students connect the solenoid to a variable power supply. A magnetic field sensor is used to measure the magnetic field strength along the central axis at the center of the solenoid. The plot of the magnetic field strength B as a function of the emf $\mathcal{E}$ of the power supply is shown below.

Figure for Question (b) — AP Physics C: Electricity & Magnetism

[ 3 ]

LO CNV-8.C.c, SP 3.D 1 point

On the cutaway view below, clearly draw an Amperian loop that can be used to determine the magnetic field at point P at the center of the solenoid.

Figure for Answer to Question (b) — AP Physics C: Electricity & Magnetism

For drawing a rectangle with one side along the central axis of the solenoid and another side outside the solenoid and whose edges do not extend beyond the solenoid

1 point

LO CNV-8.C.c, SP 5.A, 5.E 2 points

Use Ampere's law to derive an expression for the magnetic field strength at point P. Express your answer in terms of $I, \ell, N, a$ , and physical constants, as appropriate.

For using Ampere's law to calculate the magnetic field along the axis of the solenoid

1 point

$\int B \cdot d \ell=\mu_{0} I_{e n c} \therefore\left(\int B \cdot d \ell\right)_{1}+\left(\int B \cdot d \ell\right)_{2}+\left(\int B \cdot d \ell\right)_{3}+\left(\int B \cdot d \ell\right)_{4}=\mu_{0} I_{e n c}$ $\left(\int B \cdot d \ell\right)_{1}+0+\left(-\left(\int B \cdot d \ell\right)_{1}\right)+B h=\mu_{0} \frac{N}{\ell} h I$ $B h=\mu_{0} \frac{N}{\ell} h I$

For a correct answer

1 point

$B=\frac{\mu_{0} N I}{\ell}$
Some physics students conduct an experiment to determine the resistance $R_{S}$ of a solenoid with radius $a=0.015 \mathrm{~m}$ , total turns N=100, and total length $\ell=0.40 \mathrm{~m}$ . The students connect the solenoid to a variable power supply. A magnetic field sensor is used to measure the magnetic field strength along the central axis at the center of the solenoid. The plot of the magnetic field strength B as a function of the emf $\mathcal{E}$ of the power supply is shown below.

(c)

On the graph above, draw a best-fit line for the data.

Use the straight line to determine the resistance $R_{S}$ of the solenoid used in the experiment.

[ 3 ]

(d)

One of the students notes that the horizontal component of the magnetic field of Earth is $2.5 \times 10^{-5} \mathrm{~T}$ .

Is there evidence from the graph that the horizontal orientation of the solenoid affects the measured values for B ? Yes No
Justify your answer.

Would the horizontal orientation of the solenoid affect the calculated value for $R_{S}$ ? Yes No
Justify your answer.

Figure for Question (d) — AP Physics C: Electricity & Magnetism

A thin conducting loop of radius b and resistance $R_{L}$ is placed concentric with the solenoid, as shown above. The current in the solenoid is decreased from I to zero over time $\Delta t$ .

[ 2 ]

LO CNV-8.E.a, SP 2.D
1 point

Is there evidence from the graph that the horizontal orientation of the solenoid affects the measured values for B ? Yes No

Justify your answer.

If the line on the graph does not go through the origin, select "Yes"

For a correct justification

1 point

Example: The horizontal component of Earth's magnetic field will add or subtract from
the magnetic field of the solenoid depending on the orientation of the solenoid.

Alternate Points

Alternate Solution

If the line on the graph does pass through the origin, select "No"

For a correct justification

1 point

Example: Based on the graph, the line passes through the origin, so the magnetic field is
zero when the emf of the power supply is zero, therefore Earth's magnetic field is
not affecting the values of B.

LO CNV-8.E.a, SP 2.E
1 point

Would the horizontal orientation of the solenoid affect the calculated value for $R_{S}$ ? Yes No

Justify your answer.

Select "No"

For a correct justification

1 point

Example: The horizontal component of Earth's magnetic field will not affect the change
in magnetic field as the emf is changed. Therefore, the value for the resistance of the
solenoid will not change.

Figure for Answer to Question (d) — AP Physics C: Electricity & Magnetism

A thin conducting loop of radius b and resistance $R_{L}$ is placed concentric with the solenoid, as shown above. The current in the solenoid is decreased from I to zero over time $\Delta t$ .

[Maximum number: 1]

Two small spheres have equal and opposite charges and are travelling parallel to each other with speed v to the right, as shown above. What is the direction of the magnetic field midway between the spheres at the instant shown?

A

Out of the page

B

Into the page

C

Toward the bottom of the page

D

Toward the top of the page

E

Undefined, since the magnitude of the magnetic field is zero.

Figure for Question 12 Magnetic Fields and Electromagnetism question 2 — AP Physics C: Electricity & Magnetism

[Maximum number: 4]

E\&M.3.
A conducting bar of mass M, length L, and negligible resistance is connected to two long vertical conducting rails of negligible resistance. The two rails are connected by a resistor of resistance R at the top. The entire apparatus is located in a magnetic field of magnitude B directed into the page, as shown in the figure above. The bar is released from rest and slides without friction down the rails.

Is the magnitude of the net magnetic field above the bar at point C greater than, less than, or equal to the magnitude of the net magnetic field before the bar is released? Greater than Less than Equal to
Justify your answer.

While the bar is above point D, is the magnitude of the net magnetic field at point D greater than, less than, or equal to the magnitude of the net magnetic field before the bar is released? Greater than Less than Equal to
Justify your answer.

Express your answers to parts (c) and (d) in terms of M, L, R, B, and physical constants, as appropriate.

[Maximum number: 8]

Long, parallel wires S and T are a distance 2 d apart. Both wires carry equal currents I, but the

currents are in opposite directions. Both wires are parallel to the x-axis. At the instant shown

in Figure 1, Sphere 1 is a distance d above Wire S, Sphere 2 is a distance d below Wire S, and

both spheres are moving with speed v in the +x-direction. Each sphere has positive charge +Q.

Gravitational effects are negligible.

Figure 1

(a)

$F_{1}$ is the magnitude of the magnetic force exerted on Sphere 1 due to the currents in wires S

and T. $F_{2}$ is the magnitude of the magnetic force exerted on Sphere 2 due to the currents in

wires S and T.

Indicate whether $F_{2}$ is greater than, less than, or equal to $F_{1}$ by writing one of the following.

- $F_{2}>F_{1}$

- $F_{2}<F_{1}$

- $F_{2}=F_{1}$

Justify your answer.

[ 3 ]

\multirow[t]{4}{*}{A} & For indicating that $F_{2}>F_{1}$ & Point A1 \\ \hline & For correctly relating the magnitude of the magnetic force to the magnitude of the magnetic field & Point A2 \\ \hline & For indicating that the magnitude of the magnetic field at the location of Sphere 2 is greater than the magnitude of the magnetic field at the location of Sphere 1 by referring to either: - The direction of the magnetic fields from Wires S and T at the locations of the spheres & Point A3 \\ \hline &

Example Response
The magnitude of the magnetic force exerted on a sphere is directly proportional to Q, v, and B. Because Q and v are the same for both spheres, the difference in the magnitudes of the magnetic forces is due to the difference in the magnitudes of the magnetic fields at the locations of the spheres. The magnitude of the magnetic field is greater at the location of Sphere 2 than at the location of Sphere 1 because the fields from the wires are in the same direction. Therefore, $F_{2}>F_{1}$ .

Example Response

The magnitude of the magnetic force exerted on a sphere is directly proportional to Q, v, and B. Because Q and v are the same for both spheres, the difference in the magnitudes of the magnetic forces is due to the difference in the magnitudes of the magnetic fields at the locations of the spheres. The magnitude of the magnetic field is greater at the location of Sphere 2 than at the location of Sphere 1 because the fields from the wires are in the same direction. Therefore,

F_{2}>F_{1}

.

& \\ \hline

(b)

Derive an expression for the magnitude $B_{\text {tot }}$ of the magnetic field at the location of Sphere 2

due to the currents in wires S and T in terms of d, I, and physical constants, as appropriate.

Begin your derivation by writing a fundamental physics principle or an equation from the

reference information.

[ 3 ]

\multirow[t]{4}{*}{B} &

For a multistep derivation that includes the equation $\oint \vec{B} \cdot d \vec{\ell}=\mu_{0} I_{\text {enc }}$ or $d \vec{B}=\frac{\mu_{0}}{4 \pi} \frac{I(d \vec{\ell} \times \hat{r})}{r^{2}}$
Scoring Note: Vector notation is not required for this point to be earned. A multistep derivation that includes $B=\frac{\mu_{0} I}{2 \pi d}$ can earn this point.

& Point B1 \\ \hline & For a correct expression for the magnitude of the magnetic field due to the current in one wire at the location of Sphere 2 & Point B2 \\ \hline & For indicating that the magnitude of the total field is twice that due to one wire & Point B3 \\ \hline &

Example Response
Determine the magnitude B of the magnetic field due to one wire. $\oint \vec{B} \cdot d \vec{\ell}=\mu_{0} I_{\mathrm{enc}}$ $\begin{aligned} & B(2 \pi d)=\mu_{0} I & B=\frac{\mu_{0} I}{2 \pi d} \end{aligned}$
Determine $B_{\text {tot }}$ . $\begin{aligned} & B_{\text {tot }}=\frac{\mu_{0} I}{2 \pi d}+\frac{\mu_{0} I}{2 \pi d} & B_{\text {tot }}=\frac{\mu_{0} I}{\pi d} \end{aligned}$

& \\ \hline \end{tabular}

(c)

Later, Wire T carries current 3 I in the +x-direction. At the instant shown in Figure 2,

Sphere 2 is a distance d below Wire S and is moving with speed v in the +x-direction. $F_{\text {new }}$ is

the new magnitude of the magnetic force exerted on Sphere 2 due to the currents in wires S

and T.

Figure 2

Indicate whether $F_{\text {new }}$ is greater than, less than, or equal to $F_{2}$ by writing one of the

following.

- $F_{\text {new }}>F_{2}$

- $F_{\text {new }}<F_{2}$

- $F_{\text {new }}=F_{2}$

Briefly justify your answer by referencing your derivation in part B.

[ 2 ]

\multirow[t]{3}{*}{C} & For indicating $F_{\text {new }}=F_{2}$ & Point C1 \\ \hline &

For indicating $B_{\text {tot }}=\frac{3 \mu_{0} I}{2 \pi d}-\frac{\mu_{0} I}{2 \pi d}$
OR
For indicating that the net magnitude of the magnetic field at the location of Sphere 2 will remain the same even though the field from Wire T will change

& Point C2 \\ \hline &

Example Response
The magnetic field at the location of Sphere 2 due to the current in Wire T is in the opposite direction to the magnetic field due to the current in Wire S. The current in Wire T is increased by a factor of 3. Therefore, $B_{\text {tot }}=\frac{3 \mu_{0} I}{2 \pi d}-\frac{\mu_{0} I}{2 \pi d}=\frac{\mu_{0} I}{\pi d}$ in this scenario. Thus, $F_{\text {new }}=F_{2}$ .

& \\ \hline \end{tabular}

AP Physics C E and M Unit 12: Magnetic Fields and Electromagnetism

Covered syllabus points

Covered syllabus points

12 Magnetic Fields and Electromagnetism question 1

Question (a)

Question (b)

Question (c)

Question (d)

12 Magnetic Fields and Electromagnetism question 2

12 Magnetic Fields and Electromagnetism question 3

12 Magnetic Fields and Electromagnetism question 4

Question (a)

Question (b)

Question (c)