Electromagnetism - Capacitor

The capacitor shown in Figure 1 above is charged by connecting switch S to contact a. If switch S is thrown to contact b at time t = 0, which of the curve in Figure 2 above represents the magnitude of the current through the resistor R as a function of time?

A. A
B. B
C. C
D. D
E. E

(GR9677 #01)

Solution:

The capacitor is charged by connecting switch S to contact a, so the current after connecting switch to contact b (t = 0) must start with I = V/R.

Only plot A and B are right.

The current can not be the same all the time and since the voltage of a capacitor follows an exponential decay:

V(t) = V₀e^−t/RC 
 I(t) = V(t)/R = V₀e^−t/RC/ R

Only plot B is right.

Answer: B

Electromagnetism - Faraday’s law

The circuit shown is in a uniform magnetic field that is into the page and is decreasing in magnitude at rate of 150 tesla/second. The ammeter reads

A. 0.15 A
B. 0.35 A
C. 0.50 A
D. 0.65 A
E. 0.80 A

(GR9677 #02)

Solution:

V − IR − ɛ = 0
I = (V − ɛ)/R

ɛ = − dΦ/dt = −AdB/dt

Given:
dB/dt = −150 t/s (minus sign because it’s decreasing)
A = (0.1 m)² = 0.01 m²
R = 10 Ω
V = 5 V

ɛ = − (0.01)(−150) = 1.5 V
I = (5 − 1.5)/10 = 3.5/10 = 0.35 A

Answer: B

Electromagnetism - Electric Potential

Question 3-4: refer to a thin, nonconducting ring of radius R, as shown below, which has a charge Q uniformly spread out on it.

  

The electric potential at a point P, which is located on the axis of symmetry a distance x from the center of the ring, is given by

A. Q / (4πɛ₀x)
B. Q / [4πɛ₀(R² + x²)^1/2]
C. Qx / [4πɛ₀(R² + x²)]
D. Qx / [4πɛ₀(R² + x²)^3/2]
E. QR / [4πɛ₀(R² + x²)]

(GR9677 #03)

Solution:

Electric Potential, V = kQ/r 
with k = 1/4πɛ₀

The distance r of P from the charged ring is r² = R² + x²

V = Q / [4πɛ₀(R² + x²)^1/2]

Answer: B

Electromagnetism - Oscillation

Question 3-4: refer to a thin, nonconducting ring of radius R, as shown below, which has a charge Q uniformly spread out on it.

A small particle of mass m and charge –q is placed at point P and released. If R ≫ x, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to:

(GR9677 #04)

Solution:

F_electric = kqQ/r²
F_centripetal = mv²/r = mω²r

F_e = F_c
kqQ/r² = mω²r
ω² = kqQ/mr³

with
k = 1/4πɛ₀
r²  = R² + x²
R ≫ x
r² ∼ R²  → r³ ∼ R³

ω = √(^qQ/_4πɛ0mR³)

Answer: A

Notes: see problem GR9277 #65

Electromagnetism - Electric Force

Two identical conducting spheres, A and B, carry equal charge. They are initially separated by a distance much larger than their diameters, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B is equal to.

A. 0
B. F/16
C. F/4
D. 3F/8
E. F/2

(GR9677 #24)

Solution:

Coulomb's Law, F_i = kq_Aq_B/r² 
q_A = q_B = Q
F_i = F = kQ² /r² 

When uncharged sphere C touches A, charge A distributes itself evenly:
q_C = q_A' = ¹/₂ q_A = ¹/₂Q

When sphere C touches B:
q_C'  = q_B'  = ¹/₂(q_C + q_B) = ¹/₂(¹/₂Q + Q) = ³/₄Q

Therefore,
F_f  = kq_A' q_B'/r² 
= k (¹/₂Q) (³/₄Q)/r²
= ³/₈ kQ²/r² 
= ³/₈ F

Answer: D

Electromagnetism - Capacitor

Two real capacitors of equal capacitance (C₁ = C₂) are shown in the figure. Initially, while the switch S is open, one of the capacitors is uncharged and the other carries charge Q₀. The energy stored in the charged capacitor is U₀. Sometime after the switch is closed, the capacitors C₁ and C₂ carry charges Q₁ and Q₂, respectively; the voltages across the capacitors are V₁ and V₂; and the energies stored in the capacitors are U₁ and U₂.Which of the following statements is INCORRECT?

A. Q₀ = ½ (Q₁ + Q₂)
B. Q₁ = Q₂
C. V₁ = V₂
D. U₁ = U₂
E. U₀ = U₁ + U₂

(GR9677 #25)

Solution:

Initial charge: Q₀
Charge is conserved and distribute evenly
Q₀ = ½ (Q₁ + Q₂)
(A) CORRECT

C = Q/V
For parallel capacitor V = V₁ = V₂
Given C₁ = C₂ → Q₁ = Q₂
(B) and (C) CORRECT

U = ½QV
V₁ = V₂ and Q₁ = Q₂ 
→ U₁ = U₂
(D) CORRECT

U = ½QV → Q = 2U/V
From (A) Q₀ = ½ (Q₁ + Q₂),
2U₀/V₀ = ½ (2U₁/V₁ + 2U₂/V₂)
V₀ = V₁ = V₂ → 2U₀ = U₁ + U₂

(E) INCORRECT

Answer: E

Electromagnetism - RLC Circuit

A series RLC circuit is used in a radio to tune an FM station broadcasting at 103.7 MHz. The resistance in the circuit is 10 ohms and the inductance is 2.0 microhenries. What is the best estimate of the capacitance that should be used?

A. 200 pF
B. 50 pF
C. 1 pF
D. 0.2 pF
E. 0.02 pF

(GR9677 #26)

Solution:

X_L = X_C
ωL = 1 / ωC
C = 1 / ω²L 

Given:
f  = 103.7 × 10⁶ Hz
ω = 2πf  = 2π × 103.7 × 10⁶ ≈ 2π × 10² × 10⁶ = 2π × 10⁸
ω² = (2π × 10⁸)² ≈ 4π² × 10¹⁶ 
L  = 2.0 microhenries = 2 × 10⁻⁶ H

C = 1 / (4π² × 10¹⁶ × 2 × 10⁻⁶) 

= 1 / (8π² × 10¹⁰)

Take π² ≈ 10

C  ≈ 1/8 × 10⁻¹¹ 
= 1.25 × 10⁻¹² 
= 1.25 pF

Answer: C

Electromagnetism - Oscilloscope

The figure above represents the trace on the screen of cathode ray oscilloscope. The screen is graduated in centimeters. The spot on the screen moves horizontally with a constant speed of 0.5 centimeter/millisecond and the vertical scale is 2 volts/centimeter. The signal is a superposition of two oscillations. Which of the following are most nearly the observed amplitude and frequency of these two oscillations? 

Oscillation 1 Oscillation 2 A. 5V, 250Hz 2.5V, 1000Hz B.  1.5V, 250Hz 3V, 1500Hz  C. 5V, 6Hz 2V, 2Hz  D. 2.5V, 83Hz 1.25V, 500Hz  E. 6.14V, 98Hz                1.35V, 257Hz

(GR9677 #28)

Solution:

The graph shows one big λ consists of 6 small λs.
λ_big ≈ 6 cm
λ_small ≈ 1 cm 

v = λf

Given: v = 0.5 cm/ms
f_big = v/λ_b = 0.5 cm/(6 cm ms) = 1/ (12 ms) = 10³/(12 s) = 83 Hz (Osc. 1)
f_small = v/λ_s = 0.5 cm/(1 cm ms) = 1/ (2 ms) = 10³/(2 s) = 500 Hz (Osc. 2)

Answer: D

Electromagnetism - Stokes Theorem

The line integral of  u = yi − xj + zk around a circle of radius R in the xy-plane with center at the origin is equal to

A. 0
B. 2πR
C. 2πR²
D. πR²/4
E. 3R³

(GR9677 #43)

Solution:

Stokes' Theorem: The line integral of a vector field around a closed curve is equal to the surface integral of the curl over that vector field.

∮ F · dl = ∫ (∇ × F) · dA

Given:  u = yi − xj + zk 

∇ × u =

= 0 + 0 − k̂ − k̂ − 0 − 0 = − 2k̂

∫ (∇ × u) · dA = − ∫ 2k̂ dA = − 2k̂ ∫ dA  = − 2k̂ A  = − 2πR² k̂

Answer: C

Electromagnetism - High and Low-Pass Filters

The circuits below consist of two-element combinations of capacitors, diodes and resistors. V_in represents an ac-voltage with variable frequency. It is desired to build a circuit for which V_out ≈ V_in at high frequencies and V_out ≈ 0 at low frequencies. Which of the following circuits will perform this task?

(GR9677 #45)

Solution:

High-pass filter: V_out ≈ V_in at high frequencies, (ω → ∞)
Low-pass filter: V_out ≈ 0 at low frequencies (ω → 0)

Diode is a device to allow current flow in one direction only. A combination of Diode and Resistor will not affect the output frequency.
→ (B), (C) are FALSE

(A), (D), (E) circuits are combination of resistor and capacitor.

Capacitive reactance, X_C  = 1 / ωC
X_C → 0 for ω → ∞
X_C → ∞ for ω → 0

(A) FALSE
It is parallel combination of capacitor and resistor, voltage is uniform throughout the circuit.

(D) FALSE
It is Low-pass filter only.
Using Voltage Divider for series circuit: V_out = X_C / (R + X_C) V_in

X_C → 0 for ω → ∞

V_out = 0 / (R + 0) V_in ≈ 0

(E) TRUE
It is High and Low-pass filter.
Using Voltage Divider: V_out = R / (R + X_C) V_in

X_C → 0 for ω → ∞

V_out = R / (R + 0) V_in 
V_out ≈ V_in

X_C → ∞ for ω → 0
V_out = R / (R + ∞) V_in 
V_out ≈ 0

Answer: E

Notes: for similar problem see GR0177 #39

Electromagnetism - Faraday’s law

A circular wire loop of radius R rotates with an angular speed ω in a uniform magnetic field B, as shown in the figure. If the emf ɛ induced in the loop is ɛ₀ sin ωt, then the angular speed of the loop is

A. ɛ₀ R/B
B. 2πɛ₀/R
C. ɛ₀/BπR²
D. ɛ₀²/BR²
E. tan⁻¹(ɛ₀/Bc)

(GR9677 #46)

Solution:

Faraday's Law: ɛ = − dΦ/dt
with  Φ = NBA  

Given:
N = 1
B uniform (constant)
ɛ =  ɛ₀ sin ωt

→ ɛ = − B dA/dt =  ɛ₀ sin ωt

− B dA =  ɛ₀ sin ωt dt 

 B ∫dA =  ɛ₀ ∫ sin ωt dt 

− BπR² = − ^ɛ₀/_ω  cos ωt

ω = ɛ₀ cos ωt / BπR²

at t = 0,
ω = ɛ₀ / BπR²

Answer: C

Alternative solution:
A = πR²  cos ωt
ɛ = − dΦ/dt = − B dA/dt 
ɛ = − BπR²  d(cos ωt)/dt 
ɛ₀ sin ωt = BπR²ω  sin ωt 
ɛ₀= BπR²ω  

ω = ɛ₀ / (BπR²)

Electromagnetism - Faraday’s law

A wire is being wound around a rotating wooden cylinder of radius R. One end of the wire is connected to the axis of the cylinder, as shown in the figure. The cylinder is placed in a uniform magnetic field of magnitude B parallel to its axis and rotates at N revolutions per second. What is the potential difference between the open ends of the wire?

A. 0
B. 2πNBR
C. πNBR²
D. BR²/N
E. πNBR³

(GR9677 #47)

Solution:

Faraday's Law: ɛ = − dΦ/dt
with  Φ = NBA  

Given:
N revolution per second
B uniform (constant)
A constant

ɛ = − BA dN/dt
= − BπR²N

|ɛ| = πNBR²

Answer: C

Special Relativity - Electric Field

The infinity xy-plane is a nonconducting surface, with surface charge density σ, as measured by an observer at rest on the surface. A second observer moves with velocity v relative to the surface, at height h above it. Which of the following expressions gives the electric measured by this second observer?

A.

B.

C.

D.

E. 

(GR9677 #49)

Solution:

E_z = γE_0z 
with γ = 1/(1 − ^v2/_c²)^½ 

E_z = E₀/(1 − ^v2/_c²)^½

For infinity plane: E = σ/2ɛ₀ (show)

For infinity xy-plane: E₀  = σ / 2ɛ₀

E_z =  σ / 2ɛ₀(1 − ^v2/_c²)^½

Answer: C

Electromagnetism - Superposition

Questions 54-55 concern a plane electromagnetic wave that is a superposition of two independent orthogonal plane waves and can be written as the real part of 

 E = x̂ E₁ exp [i(kz −ωt)] +  ŷ E₂ exp [i(kz − ωt + π)]
where k, ω, E₁ and E₂ are real

If E₂ = E₁, the tip of the electric field vector will describe a trajectory that, as viewed along the z-axis from positive z and looking toward the origin, is a

A. Line at 45^o to the + x-axis
B. Line at 135^o to the + x-axis
C. Clockwise circle
D. Counterclockwise circle
E. Random path

(GR9677 #54)

Solution:

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂ e^{i(kz − ωt +π)} 
E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂ e^{i(kz − ωt)} · e^iπ 

with 
E₂ = E₁ = E
e^iπ = −1

E = E e^{i(kz − ωt)} x̂ − E e^{i(kz − ωt)} ŷ 
E = a x̂ − a ŷ 

tan θ  = a / (−a) = −1

tan 45  = 1
tan 135 = tan 315 = −1

Answer: B


Note: 
e^iϕ = cos ϕ + isin ϕ
e^iπ = cos π + isin π =  −1 + 0 = −1

Electromagnetism - Polarization

Questions 54-55 concern a plane electromagnetic wave that is a superposition of two independent orthogonal plane waves and can be written as the real part of 

E = x̂ E₁ exp [i(kz −ωt)] +  ŷ E₂ exp [i(kz − ωt + π)]

where k, ω, E₁ and E₂ are real

If the plane wave is split and recombined on a screen after the two portions, which are polarized in the x- and y- directions, have traveled an optical path difference of 2π/k, the observed average intensity will be proportional to

A. E₁² + E₂² 
B. E₁² − E₂² 
C. (E₁+ E₂)² 
D. (E₁− E₂)² 
E. 0

(GR9677 #55)

Solution:

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i(kz − ωt +π )} 

Path difference of 2π/k
E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i[k(z + 2π/k) − ωt + π)} 

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i[kz − ωt)} · e^i3π 
e^i3π = −1

E = x̂ E₁ e^{i(kz − ωt)}  −  ŷ E₂  e^{i(kz − ωt)} 

Intensity in the x- directions, I_x = |E₁|²  
Intensity in the x- directions, I_y = |−E₂|² 

I_total = I_x + I_y = E₁² + E₂²

Answer: A

Electromagnetism - Gauss’ Law, Electric Field

A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar², where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric field is:

A. A/4πɛ₀
B. AR³/40ɛ₀
C. AR³/24ɛ₀
D. AR³/5ɛ₀
E. AR³/3ɛ₀

(GR9677 #61)

Solution:

The net charge within the Gaussian surface with ρ as a function of r (non-uniformly charged sphere):

dq_enclose = ρ dV = Ar² d(⁴⁄₃ πr³) = Ar² ⁴⁄₃ π3r² dr = A4πr⁴ dr





Answer: B

Electromagnetism - Capacitor

Two capacitors of capacitance 1.0 microfarad and 2.0 microfarad are each charged by being connected across a 5.0-Volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the 2.0 microfarad capacitor?

A. 0 V
B. 0.6 V
C. 1.7 V
D. 3.3 V
E. 5.0 V

(GR9677 #62)

Solution:

The charges stored at each capacitor:

Q₁ = C₁ × V_t = 1 × 5 = 5 μF
Q₂ = C₂ × V_t = 2 × 5 = μF

The plates of opposite charge are connected together (series connection):

Q₁'  =  Q₂' = Q_t
1/C_t  = 1/C₁  +  1/C₂ = 1  +  1/2 = 3/2
C_t = 2/3
Since  V_t = 5
Q₂' = Q_t  =  C_t × V_t = 2/3 × 5 = 10/3

→ V₂ = Q₂' / C₂ = 10/3 × 1/2 =  10/6 = 1.66 V

Answer: C

Electromagnetism - Conductor

Two long conductors are arranged as shown above to form overlapping cylinders, each of radius r, whose centers are separated by a distance d. Current of density J flows into the plane of the page along the shaded part of one conductor and equal current flows out of the plane of the page along the shaded portion of the other, as shown. What are the magnitude and direction of the magnetic field at point A?

A. (µ₀ ⁄ 2π)πdJ, in the +y-direction
B. (µ₀ ⁄ 2π)d²J ⁄ r, in the +y-direction
C. (µ₀ ⁄ 2π)4d²J ⁄ r, in the −y-direction
D. (µ₀ ⁄ 2π)J²r ⁄ d, in the −y-direction
E. There is no magnetic field at A

(GR9677 #69)

Solution:

Using right-hand rule, the B direction is in the +y direction
→ C, D, E are FALSE

For option (B), if r → 0,  B → ∞, B cannot be infinite.

Answer: A

Electromagnetism - EM Radiation

A charged particle, A, moving at a speed much less than c, decelerates uniformly. A second particle, B, has one-half the mass, twice the charge, three times the velocity, and four times the acceleration of particle A. According to classical electrodynamics, the ratio P_B/P_A of the powers radiated is

A. 16
B. 32
C. 48
D. 64
E. 72

(GR9677 #70)

Solution:

Larmor formula:
Power radiated by an accelerating non-relativistic point charge particle,
P = q²a² ⁄ 6πɛ₀c³

P does not depend on m and v.

P_B/P_A =  (2q)² (4a)² ⁄q²a²  =  64

Answer: D

Electromagnetism - Particle Trajectory

The figure above shows the trajectory of a particle that is deflected as it moves through the uniform electric field between parallel plates. There is potential difference V and distance d between the plates, and they have length L. The particle (mass m, charge q) has non relativistic speed v before it enters the field, and its direction at this time is perpendicular to the field. For small deflections, which of the following expressions is the best approximation to the deflection angle θ?

A. Arctan ( (L/d)(Vq/mv²) )
B. Arctan ( (L/d)(Vq/mv²)² )
C. Arctan ( (L/d)² (Vq/mv²) )
D. Arctan ( (L/d)(2Vq/mv²)^½ )
E. Arctan ( (L/d)^½(2Vq/mv²) )

(GR9677 #71)

Solution:

tan θ = v_y / v_x = at / v 
v  = L/t → t = L/v
F  =  ma = qE = qV / d → a = qV / md

tan θ = at / v =  (qV / md) (L/v) / v = (qVL /mdv²)
→  θ = arctan (qVL /mdv²) = arctan ( (L/d)(Vq/mv²) )

Answer: A