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 - Conductor

A coaxial cable having radii a, b, and c carries equal and opposite currents of magnitude i on the inner and outer conductors. What is the magnitude of the magnetic induction at point P outside of the cable at a distance r from the axis?

A. Zero

B.

C.

D.

E.

(GR9277 #09)

Solution:

The inner and outer conductors carry equal and opposite currents. Thus, the magnitude of the magnetic induction outside the coaxial cable is zero.

Answer: A 

Electromagnetism - Ampere's Law

Questions 54-55

A rectangular loop of wire with dimensions shown above is coplanar with a long wire carrying current I. The distance between the wire and the left side of the loop is r. The loop is pulled to the right as indicated. 

What is the magnitude of the net force on the loop when the induced current is i?

A. 

B. 

C. 

D. 

E. 

(GR9277 #55)

Solution:

Magnetic force on an electric current: 

For the loop, induced current I = i is parallel to B, and  l = b

 → F = ibB
Ampere's Law for a magnetic field around a straight wire: 
→  

Only choice D fits the above equation.

Answer: D


Complete calculation:

On the left, r  → r
On the right, r  → r + a






Note: since force on the left is directed to the left → negative sign

The magnitude is 

Electromagnetism - Faraday's Law

A uniform and constant magnetic field B is directed perpendicularly into the plane of the page everywhere within a rectangular region as shown above. A wire circuit in the shape of a semicircle is uniformly rotated counterclockwise in the plane of the page about an axis A. The axis A is perpendicular to the page at the edge of the field and directed through the center of the straight-line portion of the circuit. Which of the following graphs best approximates the emf ε induced in the circuit as a function of time t?

(GR9277 #57)

Solution:

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

Given:
N = 1,
B constant
A increases and decreases uniformly since the wire circuit rotates uniformly
→  the rate of change of A, dA/dt = constant.

ɛ = − BdA/dt  = constant

Only graph (A) shows constant ε.
ε changes periodically from positive to negative since only half of area covered in magnetic field.

Answer: A

Electromagnetism - Method of Image

A positive charge Q is located at a distance L above an infinite grounded conducting plane, as shown in the figure above. What is the total charge induced on the plane?

A. 2Q
B. Q
C. 0
D. –Q
E. –2Q

(GR0177 #08)

Solution:

Method of image:

• +Q attracts negative charges in the plate and repels positive ones. 
• As the positive charges want to “get away”, they succeed in doing so through the ground, leaving the negative charges behind. 
• The plate is left with a net negative charge.

Answer: D

Electromagnetism - RLC Circuit

An AC circuit consists of the elements shown above, with R = 10,000 ohms, L = 25 millihenrie and C an adjustable capacitance. The AC voltage generator supplies a signal with amplitude of 40 volts and angular frequency of 1,000 radians per second. For what value of C is the amplitude of the current maximized?

A. 4 nF
B. 40 nF
C. 4 μF
D. 40 μF
E. 400 μF

(GR0177 #38)

Solution:

I_max when X_L = X_C

→ ωL = 1 / ωC

C = 1 / ω²L 
= 1 / [(10³)² × 25 × 10⁻³]
= 1 / (25 × 10³)
= 40 × 10⁻⁶ 
= 40 μF

Answer: D

Electromagnetism - High-Pass Filters

Which two of the following circuits are high-pass filters

A. I and II
B. I and III
C. I and IV
D. II and III
E. II and IV

(GR0177 #39)

Solution:

High-pass filter: V_out ≈ V_in  for frequency, ω → ∞

Inductive reactance, X_L = ωL
Capacitive reactance, X_C  = 1 / ωC

For ω → ∞
X_L → ∞
X_C → 0

Voltage divider:


Case I.

→ Low-pass filter


Case II.

Note: R ≪ ∞
→ High-pass filter


Case III.

→ High-pass filter


Case IV.

→ Low-pass filter

Answer: D

Electromagnetism - RL Circuit

 

In the circuit shown above, the switch S is closed at t = 0. Which of the following best represents the voltage across the inductor, as seen on an oscilloscope?

(GR0177 #40)

Solution:

The voltage drops across the resistor and the inductor → (A), (B), (C) are FALSE.

The time constant when the voltage decreases slowly is



Answer: D

Electromagnetism - LC circuit

For an inductor and capacitor connected in series, the equation describing the motion of charge is

where L is the inductance, C is capacitance, and Q is the charge. An analogous equation can be written for a simple harmonic oscillator with position x, mass m, and spring constant k. Which of the following correctly lists the mechanical analogs of L, C, and Q? 

	L	C	Q
A	m	k	x
B.	m	1/k	x
C.	k	x	m
D.	1/k	1/m	x
E.	x	1/k	1/m

(GR0177 #59)

Solution:

The form of SHO:

LC circuit equation: 

L = m
C = 1/k
Q = x

Answer: B

Electromagnetism - Faraday’s law

A coil of 15 turns, each of radius 1 centimeter, is rotating at a constant angular velocity ω = 300 radians per second in an uniform magnetic field of 0.5 Tesla, as shown in figure. Assume at time t = 0 that the normal n̂ to the coil plane is along the y-direction and that the self-inductance of the coil can be neglected. If the coil resistance is 9 ohms, what will be the magnitude of the induced current in milliamperes?

A. 225π sin ωt
B. 250π sin ωt
C. 0.08π cos ωt
D. 1.7π cos ωt
E. 25π cos ωt

(GR0177 #86)

Solution:

Ohm’s Law: ɛ = V = IR
→  I = ɛ / R

Faraday's Law: ɛ = − dΦ_B / dt

Magnetic Flux:  dΦ_B =  NB_⊥dA

At time t = 0, the normal to the coil plane is along the y-direction.

It means: n̂ ∥ ŷ → B ∥ A → Φ_B = 0

→ Φ_B =  NBA sin ωt, since Φ_B (t = 0) = 0

Φ_B = − NB₀ πr² sin ωt
ɛ = − dΦ_B / dt = NB₀ πr² ω cos ωt

I =  ɛ / R = (NB₀ωr²/ R) π cos ωt

N  = 15 turns
B₀ = 0.5 Tesla
ω = 300 rad/s
r  = 1 cm =  10⁻²  m
R = 9 ohms

I = (15 ×  0.5 × 300 × 10⁻⁴ / 9) π cos ωt
= (25 × 10⁻³) π sin ωt Ampere
= 25 π sin ωt milliAmpere

Answer: E

Electromagnetism - RL Circuit

In the circuit shown above, R₂ = 3R₁ and the battery of emf ξ has negligible internal resistance. The resistance of the diode when it allows current to pass through it is also negligible. At time t = 0, the switch S is closed and the currents and voltage are allowed to reach their asymptotic values. Then at time t₁, the switch is opened. Which of the following curves most nearly represents the potential at point A as a function of time t?

(GR8677 #94)

Solution:

When the switch is closed, at time t = 0, the voltage drops through the circuit. Thus, C, D and E are FALSE.

The time constant in the first process:


The time constant, τ  is the time it takes the voltage across the component to either decrease or increase.

When the switch is opened, at time t₁, the voltage increases with the time constant:

Thus,

It means, it takes less time for the voltage to increase than decrease.

Answer: B