AP® Physics C: E&M - Magnetic Flux - FRQs
Question
3. A wire is connected to a resistor of resistance R to form a rigid rectangular loop of width L and height 2L. An external force is exerted on the loop so that the loop always moves with constant speed v in the +x-direction, as shown in Figure 1. The loop then enters Region 1 of external uniform magnetic field of magnitude B that is directed in the −z-direction. Region 1 has boundaries x = L and x = 2.5L. The loop later enters Region 2 with two external, uniform magnetic fields, each of magnitude B, that are parallel but are directed in opposite z-directions. Region 2 has boundaries x = 2.5L and x = 3.5L. Point S is the midpoint of the leading edge of the loop and is aligned with the horizontal boundary in Region 2 that separates the two magnetic fields.
(a) On the following axes, sketch a graph of the magnetic flux Φ through the rectangular loop as a function of the position x of Point S from x = 0 to x = 4.5L. The +z-direction indicated in Figure 1 corresponds to +Φ.
(b) Consider the instant when Point S reaches x = 1.5L.
i. Indicate whether the current \(I_{R}\) that is induced in the rectangular loop when Point S reaches x = 1.5L is clockwise, counterclockwise, or zero.
_____ Clockwise _____ Counterclockwise _____ Zero
Briefly justify your answer.
ii. Derive an expression for \(I_{R}\) when Point S reaches x = 1.5L. If \(I_{R}\) = 0, indicate how the derived expression shows that \(I_{R}\) = 0. Express your answer in terms of R, L, v, B, and physical constants, as appropriate.
iii. Derive an expression for the power P dissipated by the resistor when Point S reaches x = 1.5L. Express your answer in terms of R, L, v, B, and physical constants, as appropriate.
The total energy dissipated by the resistor in the rectangular loop as Point S moves from x = 0 to x = 4.5L is \(E_{original}\).
The vertical boundary between regions 1 and 2 is now shifted to x = 1.5L. After the boundary is shifted, the rectangular loop again moves with speed v in the +x-direction, as shown in Figure 2. The total energy dissipated by the resistor as Point S moves from x = 0 to x = 4.5L is \(E_{new}\).
(c) Indicate whether \(E_{new}\).is greater than, less than, or equal to \(E_{original}\).
_____ \(E_{new}\). > \(E_{original}\)._____ \(E_{new}\). < \(E_{original}\). _____ \(E_{new}\). = \(E_{original}\).
Briefly justify your answer.
The original magnetic fields are modified so that the region L < < x 3.5L contains an external uniform magnetic field of magnitude B that is directed in the −z-direction. A new wire is connected to a resistor of resistance R to form a rigid triangular loop with base length L and height 2L. An external force is exerted on the loop so that the loop always moves with speed v in the +x-direction, as shown in Figure 3. Point S represents the lower-leading corner of the loop.
(d) On the following axes, sketch a graph of the induced current IT in the triangular loop as Point S moves from x = L to x = 3L.
▶️Answer/Explanation
3(a) Example Response 
Scoring Note: A sketch that is reflected across the horizontal axis can earn the four points.
Scoring Note: The absolute values of the slopes in regions L < 2D and 2.5 < x < 3.5 L are not considered for earning points in the response in part (a).
3(b)(i) Example Response
Counterclockwise. The magnetic flux through the loop is increasing in the −z-direction. Therefore, a magnetic field is induced by the current in the loop to oppose the increasing magnetic flux. To establish this field, the current must be counterclockwise.
3(b)(ii) Example Solution 
(b)(iii) Example Solution
\(p = I^{2}R\)
\(p = \frac{\left ( 2BLv \right )^{2}}{R}\)
\(p =\frac{2B^{2}L^{2}v^{2}}{R}\)
(c) Example Response
The change in magnetic flux is greater in the original scenario, which produces an emf and current for a longer time. Therefore, \(E_{new}\) < \(E_{original}\)
(d) Example Response 
Scoring Note: A response that is reflected across the horizontal axis earns both points.
Scoring Note: Any portion of the graph before x L = and after x = 3L are not scored.
Question
3. A wire is connected to a resistor of resistance R to form a rigid square loop of side length D. An external force is exerted on the loop so that the loop always moves with constant speed v in the +x direction, as shown in Figure 1. The loop then enters Region 1 of external uniform magnetic field of magnitude B that is directed in the +z-direction. Region 1 has boundaries x = D and x = 2D. The loop later enters Region 2 of external uniform magnetic field of magnitude 2B that is directed in the +z-direction. Region 2 has boundaries x = 2D and x = 3.5D. Point S is the midpoint of the leading edge of the loop.
(a) On the following axes, sketch a graph of the magnetic flux Φ through the square loop as a function of the position x of Point S from x = 0 to x = 4.5D. The +z-direction indicated in Figure 1 corresponds to +Φ.
(b)Consider the instant when Point S reaches x = 1.5D.
i. Indicate whether the current IS that is induced in the square loop when Point S reaches x = 1.5D is clockwise, counterclockwise, or zero.
_____ Clockwise _____ Counterclockwise _____ Zero
Briefly justify your answer.
ii. Derive an expression for \(I_{S}\) when Point S reaches x = 1.5L. If \(I_{S}\) = 0, indicate how the derived expression shows that \(I_{S}\) = 0. Express your answer in terms of R, L, v, B, and physical constants, as appropriate.
iii. Derive an expression for the power P dissipated by the resistor when Point S reaches x = 1.5D. Express your answer in terms of R, D, v, B, and physical constants, as appropriate.
The total energy dissipated by the resistor in the square loop as Point S moves from x = 0 to x = 4.5D is \(E_{original}\).
The vertical boundary between regions 1 and 2 is now shifted to x = 2.5D. After the boundary is shifted, the square loop again moves with speed v in the +x-direction, as shown in Figure 2. The total energy dissipated by the resistor as Point S moves from x = 0 to x = 4.5D is is \(E_{new}\).
(c) Indicate whether \(E_{new}\).is greater than, less than, or equal to \(E_{original}\).
_____ \(E_{new}\). > \(E_{original}\)._____ \(E_{new}\). < \(E_{original}\). _____ \(E_{new}\). = \(E_{original}\).
Briefly justify your answer.
The original magnetic fields are modified so that the region D < < x 3.5D contains an external uniform magnetic field of magnitude B that is directed in the +z-direction.
A new wire is connected to a resistor of resistance R to form a rigid triangular loop with base length D and height D. An external force is exerted on the loop so that the loop always moves with speed v in the +x-direction, as shown in Figure 3. Point S represents the upper-leading corner of the loop.
(d) On the following axes, sketch a graph of the induced current \(I_{T}\) in the loop as Point S moves from x = D to x = 3D.
▶️Answer/Explanation
3(a) Example Response
Scoring Note: A response that is reflected across the horizontal axis can earn the four points.
Scoring Note: The absolute values of the slopes in the entire regions D x < < 3D and 3.5D x < < 4.5D are not considered for earning these points.
3(b)(i) Example Response
Clockwise. The magnetic flux through the loop is increasing in the +z -direction. Therefore, a magnetic field is induced by the current in the loop to oppose the increasing magnetic flux. To establish this field, the current must be clockwise.
3(b)(ii) Example Solution
(b)(iii) Example Solution
\(p = I^{2}R\)
\( p = \left ( -\frac{BDv}{R} \right )^{2}\)
\(p =\frac{2B^{2}L^{2}v^{2}}{R}\)
(c) Example Response
The change in magnetic flux is greater in the original scenario, which produces an emf and current for a longer time. Therefore, \(E_{new}\) < \(E_{original}\)
(d) Example Response 
Scoring Note: A response that is reflected across the horizontal axis earns both points.
Scoring Note: Any portion of the graph before x D = and after x = 3D will not be scored.
Question
A circular wire loop with radius 0.10 m and resistance 50 Ω is held in place horizontally in a magnetic field \(\overrightarrow{B}\) directed upward at an angle of 60° with the vertical, as shown in the figure above. The magnetic field in the direction shown is given as a function of time t by B(t) = a(1- bt), where a = 4.0 T and b = 0.20 s-1 .
(a) Derive an expression for the magnetic flux through the loop as a function of time t.
(b) Calculate the numerical value of the induced emf in the loop.
(c)
i. Calculate the numerical value of the induced current in the loop.
ii. What is the direction of the induced current in the loop as viewed from point P ?
____ Clockwise ____ Counterclockwise
Justify your answer.
(d) Assuming the loop stays in its current position, calculate the energy dissipated in the loop in 4.0 seconds.
(e) Indicate whether the net magnetic force and net magnetic torque on the loop are zero or nonzero while the loop is in the magnetic field.
Net magnetic force: ____ Zero ____ Nonzero
Net magnetic torque: ____ Zero ____ Nonzero
Justify both of your answers.
Answer/Explanation
Ans:
(a)
Φm = ∫B.dA = (a) (1-bt) (πr2) (cos 600)
= \(\frac{1}{2}a(1-bt)(\pi )(0.01)=\frac{a\pi (1-bt)}{200}\)
(b)
\(\varepsilon =\frac{-d\phi m}{dt}=\frac{(-b)(a\pi )}{200}=\frac{ba\pi }{200}=\frac{0.2(4)\pi }{200}=12.6mV\)
(c) (i)
\(\varepsilon = IR\)
\(I = \frac{\varepsilon }{R}= \frac{0.0126V}{50\Omega }=0.251 mA\)
(ii)
ii. What is the direction of the induced current in the loop as viewed from point P ?
____ Clockwise ___×_ Counterclockwise
B is decreasing based on its function. By Jeny’s law the emp opposes the change in flux. By the right- hand rule the current flows counterclockwise to induce a magnetic field out of the page, in the directions of B.
(d)
\(E = Pt = \left ( \varepsilon ^{2}/R \right )t = \frac{.0126^{2}}{50}(4)= 1.27\times 10^{-5}J\)
(e) Indicate whether the net magnetic force and net magnetic torque on the loop are zero or nonzero while the loop is in the magnetic field.
Net magnetic force: __×__ Zero ____ Nonzero
Net magnetic torque: ____ Zero __×__ Nonzero
F = Il × B
By the right hand rule, the direction of induced current results in an upward-left force on the left side of the loop and a down ward-right force of equal magnitude on the right side. The sum of these forces is zero but they rotate the loop around its arise by providing a tongue.
Question
A uniform magnetic field B is directed into the page, and exists in a circular region of radius d. A single loop of wire of radius D is placed concentrically around the magnetic field region in the
plane of the page. The initial magnetic field strength is B0 • Calculate the following in terms of given values and fundamental constants.
(a) Determine the initial flux through the loop of wire. At time t = o s, the magnetic field strength as a function of time t is given by the equation B(t) = B0 t2, where B0 is a positive constant.
(b) Determine the magnitude of the induced emf in the single loop.
(c) Determine the direction of the induced current in the loop. The loop of wire has a resistance R.
(d) Determine the energy dissipated in the loop up until a given time t1
Answer/Explanation
(a) The magnetic flux is the dot product of the magnetic field and area vector that the magnetic field goes through. For this problem the area goes out to a radius d because the field
does not go out to the radius of the loop.
\(\Phi _{B}=B\cdot A\)
\(\Phi _{B}=B_{o}\cdot \pi d^{2}\)
(b) Use Faraday’s law to determine the induced emf. We will ignore the negative sign in the final answer because we only need the magnitude of the emf.
\(\varepsilon =-\frac{d\Phi _{B}}{dt}\)
\(\varepsilon =-\pi d^{2}\frac{d}{dt}(B_{o}t^{2})\)
\(\varepsilon =-\pi d^{2}(2B_{o}t)\)
\(\varepsilon =-2B_{o}\pi d^{2}t\)
\(|\varepsilon |=2B_{o}\pi d^{2}t\)
(c) Lenz’s law states that the induced emf will produce a magnetic field to oppose the change in the external magnetic field. The external magnetic field going into the page is
increasing with time, so to oppose the change in the external magnetic field, the induced current in the loop will create a magnetic field out of the page. To do this, the induced
current must be counterclockwise.
(d) Power is the rate at which energy is dissipated. Power for a circuit 1s given by the equations \(P=IV=\frac{V^{2}}{R}=I^{2}\). We will use
\(\frac{V^{2}}{R}\) for power since we know the voltage and the resistance.
\(\frac{V^{2}}{R}dt=dE\)
\(\int_{0}^{t_{1}}\frac{V^{2}}{R}dt=E\)
\(\int_{0}^{t_{1}}\frac{(2B_{o}\pi d^{2}t)^{2}}{R}dt=E\)
\(\frac{4B_{0}^{2}\pi ^{2}d^{4}t_{1}^{3}}{3R}=E\)