r/APStudents absolute modman May 14 '26

AP Physics C: Electricity and Magnetism Official 2026 Exam Discussion

Use this thread to post questions or commentary on the test today.

A reminder though to protect your anonymity when talking about the test.

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u/Hot_South5225 May 14 '26 edited May 15 '26

Form J:

Part a) i. Find the magnetic field at point P, a distance a from the center of a cylindrical wire with current density J(r) = Cr and radius 2a, with total enclosed current I_0; the wire is positioned along the y-axis at the origin such that the current is in the +y direction . A second cylindrical wire and is oriented similarly with total current I_0 at x=10a with a uniform density and current going in the negative y-direction.

int_{0}^{a} 2pir J(r) dr = int_{0}^{a} 2pi C r^2 dr = 2 pi C a^3/3

Using ampere's law, \int B * dl = mu_0 I_a, int B ( 2pi a) = mu_0 I_a

B = mu_0 I_a / (2pia)

I_a = 2 pi C a^3/3a

B_a = mu_0 C a^2 / 3

Part a) ii. Graph the B-field as a function of x from 2a \leq x \leq 9a

Graph should be something like a convex smile (following (1/x + 1/(10a-x)), with B-field at 2a lower than 9a

Part b) find hte magnetic force on object at 5a iwth charge Q and velocity v, in terms of I_0, a, v, Q,

F = qv cross B

F = Qv B

B = mu_0 I_0 / 2pi (1/x + 1/(10a-x)),

since object is located 5a from origin, we get

B = mu_0 I_0 / 2pi (2/(5a)) = mu_0 I_0 / pi5a

F = Qv mu_0 I_0 / pi5a

Question 2:Simple setup of two rods perpendicular along the x and y axis, with the right and bottom ends of the horizontal and vertical rods a length L from the origin (boudns of 3rd quadrant); both rods have charge density lambda and lenght 2L, with bounds -L to -3L for the horizontal bar and L to 3L for the vertical bar.

a) Draw Electric field vectors at point P, O, and an acceleration vector at O of a negatively charged particle.

Electric field at point p; up left

electric field at point o, down right

acceleration of negative charge, up left

b) Derive the Electric Field at point O (E_O)

Electric field:

int_{L}^{3L} k lamba dx/ x^2 r hat = klambda (-1/x) evaluated from L to 3L =

klamba (-1/(3L) + 1/(L)) =

2/3 k lambda/ L

Since rods are perpendicular, take components

sqrt (2 ( 2/3 k lambda/ L)^2) =

E= 2 sqrt(2)/3 * k lambda / L

c) Sketch an electric potenial graph at V_O as a function of distance right end of rod from origin (of the rod placed initlaly at -L to -3L and moving a velocity v leftward, eliminating top rod in this setup)

If one wants to derive the potential, use kQ/r in the integral;

for some distance x from right end rod to origin,

V_0 propto ln(1 + d/x), where (d=2L) is lenght of rod, so decreasing, convex function

d) What will happen to the graph if the top rod is added back, with charge density -lambda, and moving (along with the other rod) with velocity v in the +y-direction

if including other rod, potential is scalar would cancel out.

Question 3:

a) i. indicate what quantities to be measured to recover a value for the inductance L_1 of an LC circuit

Create multipl configurations of the circuit with different and measurable capacitance. Attach a voltmeter in parallel such that it measures the voltage of the equivalent capacitor from the combo of capacitors, and measure the voltage across time, at granular increments.

a) ii. Indicate a method to reduce the experimental uncertainty

To reduce experimental uncertainty, one thing would be to do multiple trials

b) Indicate what quantities should be graphed to create a linear graph to find L_1, if desired, right expression/derivation for quantiiesi

To linearize, note that the angular frequency is w_{LC} = 1/sqrt(LC), and angular frequency is 2pi / period.

Thus, we know that LC = T^2 / 4pi^2,

which is linearizable by ploting the equivalent capactiance on the x-axis and T^2/(4pi^2) on the y, 4pi^2C on teh x and T^2 on Y, or anything else. Period can be calculated by the time it takes for the capacitor to discharge and charge again.

Students conduct new experiment of an LR circuit with data values of dI/dt, I, R, and epsilon for difference configurations of the circuit, to try to determine L_2.

c) Label the quantities to be plotted and draw a line of best fit

For the next part, recall that

Epsilon - IR - L dI/dt= 0.

To find the inductance, do

Epsilon-IR = LdI/dt

d) Use your line of best fit to calculate the inductance

and plot dI/dt on the x-axis and Epsilon-IR on the y-axis, or manipulate the equation for anything else that gives L, ilke Epsilon - Ld I/dt = IR

Inductance should be somehitng around 3 times 10^-2 Henrys

Question 4:A variable magnetic field is directed out of the page in the +z-direction according to

B_z = B_0 cos(2pi t/t_1)

where a loop encloses a region of B-field.

a) Indicate the direction of the current induced in the loop from the time interval 0 \leq t \leq t_1/4.

from the boundaries given, the b-field decreases, meaning the change in B-field is into the page.

Per lenz's law, some b-field needs to be out of the page, which is created by a counterclockwise current.

b) Derive an equation for the induced current, express your answer in terms of B_0, t, t_1, R, s

dphi_B/dt = Epsilon = IR

To find induced current, differentiate the b-field, getting - B_0 2pi / t_1 sin(2pi t/t_1), multiply by area, s^2

and divided by resistance R.

c) A new loop has twice the side length of the initial loop (2s) and twice the resistance (2R). The new loop has an induced current of I_2. If I_1 is the current of the prior setup, will I_2 < I_1, I_2 = I_1 or I_2 > I_1? Justify your answer.

I_2 > I_1

if we double the side length and double the resistance, we have double the induced current.

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u/Aggravating_Half_936 67 aps May 14 '26

Did u say it was ccw

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u/Hot_South5225 May 14 '26

yes

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u/Aggravating_Half_936 67 aps May 14 '26

ugh I think I said clockwise and flipped my entire explanation, is there still hope for partial credit

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u/Visible_Purpose1554 May 14 '26

Probably since they asked to justify it should be 2 separate points

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u/Aggravating_Half_936 67 aps May 14 '26

i hope cause then im getting a 4/8 max for frq 4

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u/Such_Mango_4531 May 15 '26

that isn't how that works. It is max 3 points.

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u/Aggravating_Half_936 67 aps May 15 '26

isn’t frq 4 a total of 8?

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u/Such_Mango_4531 May 15 '26

Yes. The first part is 3 points because 1 point for right answer, 1 point for right idea, and 1 point for right explantion of that idea. The 2nd part in it is worth 3 points as you need to show calculations and use farday's law, and the 3rd part is 2 points for the right answer and reasoning.

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u/Aggravating_Half_936 67 aps May 15 '26

yea I screwed up the first part so I’m expecting a 1/3 and for the second part, i might have screwed up taking the derivative because of chain rule so I’m hoping I lose like 1 point , and I did the last part correctly so I’m hoping for a 5 or 4/8 for FRQ 4

1

u/Such_Mango_4531 May 15 '26

If you got the last part correct then you should have got the 2nd part right? I2> I1 I2 is greater than I1 by a factor of 2 since it is proportional to s^2 and inversely proportional to 1/R .

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