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AP Physics C: Electricity and Magnetism Medium practice

15 questions

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Medium practice

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Question 1

An RL circuit ( $R = 20 Ω$ , $L = 0.40 H$ ) is powered by $E = 10 V$ . After a long time the switch is opened. The current through the inductor at time $t$ after opening is:

I (t) = I_{0} e^{- R t / L} = 0.50 e^{- 50 t} A

I (t) = I_{0} (1 - e^{- R t / L})

I (t) = I_{0} e^{- t / R C}

(RC decay formula)D.

I (t) = 0

immediately after the switch opens.

Question 2

The energy stored in an inductor of inductance $L = 50 mH$ carrying current $I = 4 A$ is:

U_{L} = \frac{1}{2} L I^{2} = 0.40 J

U_{L} = L I^{2} = 0.80 J

U_{L} = \frac{L}{2 I ^{2}} = 0.0016 J

U_{L} = \frac{1}{2} \frac{I ^{2}}{L} = 160 J

Question 3

A resistor $R = 50 Ω$ dissipates power $P = 8 W$ . The current through it and the voltage across it are:

I = 0.4 A

;

V = 20 V

I = 4 A

;

V = 200 V

I = 0.04 A

;

V = 2 V

I = 0.16 A

;

V = 8 V

Question 4

An RC circuit consists of a resistor $R = 200 k Ω$ and capacitor $C = 50 μ F$ connected in series with a battery $V_{0} = 10 V$ . The switch is closed at $t = 0$ with the capacitor initially uncharged. The voltage across the capacitor at time $t = τ = R C$ is:

V_{C} (τ) = V_{0} (1 - e^{- 1}) \approx 6.32 V

V_{C} (τ) = V_{0} / e \approx 3.68 V

V_{C} (τ) = V_{0} = 10 V

V_{C} (τ) = V_{0} /2 = 5 V

Question 5

A capacitor of capacitance $C$ is fully charged to voltage $V_{0}$ and then disconnected from the battery. A dielectric of dielectric constant $κ$ is then inserted, filling the gap completely. What happens to the stored energy?

A.The energy decreases by a factor of

κ

.B.The energy increases by a factor of

κ

.C.The energy remains the same, because charge is conserved.D.The energy increases by a factor of

κ^{2}

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Question 6

A fully charged capacitor $C = 100 μ F$ at voltage $V_{0} = 50 V$ is discharged through resistor $R = 500 Ω$ starting at $t = 0$ . The charge on the capacitor as a function of time is $Q (t) = Q_{0} e^{- t / τ}$ . The current through the resistor at $t = 0$ is:

I (0) = V_{0} / R = 0.10 A

I (0) = Q_{0} / R C = 0.10 A

(same value, but a different derivation — both are valid).C.

I (0) = 0

, because the current builds up from zero.D.

I (0) = V_{0} / (R + 1/ C) \approx 0.10 A

Question 7

Two coils have mutual inductance $M = 0.020 H$ . Coil 1 carries a current that changes at rate $d I_{1} / d t = 30 A/s$ . The magnitude of the EMF induced in coil 2 is:

∣ E_{2} ∣ = M ∣ d I_{1} / d t ∣ = 0.60 V

∣ E_{2} ∣ = M ∣ I_{1} ∣ =

unknown (need

I_{1}

, not

d I_{1} / d t

).C.

∣ E_{2} ∣ = M^{2} ∣ d I_{1} / d t ∣ = 0.012 V

∣ E_{2} ∣ = (d I_{1} / d t) / M = 1500 V

Question 8

A circular loop of resistance $R = 5 Ω$ and radius $r = 0.10 m$ is pulled at constant velocity $v = 2 m/s$ out of a region of uniform magnetic field $B = 0.4 T$ (perpendicular to the loop). While the loop is partially out, the induced current is:

I_{ind} = B Lv / R

, where

L = 2 r = 0.20 m

is the chord in the field.B.

I_{ind} = B Lv / R

where

L

is the length of the loop edge cutting through the field boundary.C.

I_{ind} = B π r^{2} v / R

I_{ind} = 0

, because the flux is decreasing symmetrically.

Question 9

The electric potential in a region is $V (x, y) = 3 x^{2} y - y^{3}$ (in volts, with $x, y$ in meters). The $x$ -component of the electric field at the point $(1, 1)$ is:

E_{x} = - 6 V/m

E_{x} = + 6 V/m

E_{x} = - 3 V/m

E_{x} = + 3 V/m

Question 10

A charged isolated conductor has a sharp point (small radius of curvature $r$ ) and a flat region (large radius of curvature). Which statement correctly describes the charge distribution?

A.Surface charge density and electric field are highest at the sharp point.B.Surface charge density is uniform across the entire conductor surface.C.The electric field is stronger over flat regions because there is more area for charge to spread.D.Electric field inside the conductor is largest near the sharp point.

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Question 11

Using the Biot-Savart law, the magnetic field at a point $P$ on the axis of a circular loop of radius $R$ at axial distance $z$ from the center is:

B = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + z ^{2} ) ^{3/2}}

B = \frac{μ _{0} I}{2 R ^{2} + z ^{2}}

B = \frac{μ _{0} I R}{2 ( R ^{2} + z ^{2} )}

B = \frac{μ _{0} I}{2 R}

(same as at the center, independent of

z

Question 12

A parallel-plate capacitor with plate area $A = 0.10 m^{2}$ and gap $d = 2 mm$ is filled with a dielectric of $κ = 5$ . A voltage $V = 100 V$ is applied. The surface charge density $σ_{f}$ on the capacitor plates (free charge) is:

σ_{f} = κ ϵ_{0} V / d \approx 2.21 μ C/m^{2}

σ_{f} = ϵ_{0} V / d \approx 0.44 μ C/m^{2}

σ_{f} = ϵ_{0} V / (κ d) \approx 0.089 μ C/m^{2}

σ_{f} = κ^{2} ϵ_{0} V / d \approx 11.1 μ C/m^{2}

Question 13

Using Ampère's law, the magnetic field at a distance $r$ inside a long cylindrical conductor of radius $R$ carrying a total current $I$ uniformly distributed over its cross section is:

B = \frac{μ _{0} I r}{2 π R ^{2}}

for

r < R

B = \frac{μ _{0} I}{2 π r}

for

r < R

B = 0

for

r < R

B = \frac{μ _{0} I}{2 π R ^{2}}

for

r < R

Question 14

An infinite plane of charge has surface charge density $σ$ . Using Gauss's law with a pillbox Gaussian surface that straddles the sheet, the electric field magnitude on each side of the sheet is:

E = \frac{σ}{ϵ _{0}}

E = \frac{σ}{2 ϵ _{0}}

E = \frac{2 σ}{ϵ _{0}}

E = \frac{σ}{4 π ϵ _{0}}

Question 15

A particle of mass $m$ , charge $q$ , enters a uniform magnetic field $B$ perpendicular to its velocity $v$ . The radius of the circular orbit is:

r = \frac{m v}{q B}

r = \frac{q B}{m v}

r = \frac{m v ^{2}}{q B}

r = \frac{m}{q v B}

AP Physics C: Electricity and Magnetism Medium practice — Free with Answer Explanations | Test Practice Hub