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 - Magnetic Field

A coaxial cable has the cross section shown in the figure above. The shaded region is insulated. The regions in which r  a and b  r  c are conducting. A uniform dc current density of total current I flows along the inner part of the cable (r  a) and returns along the outer part of the cable (b  r  c) in the direction shown. The radial dependence of the magnitude of the magnetic field, H, is shown by which of the following?

 

(GR9677 #88)

Solution:

For r = 0 → H = 0
→ D and E are FALSE

For r = c → H = 0 since I_in and I_out cancel each other
→ A and C are FALSE

Answer:  B

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 - Method of Image

Two positive charged of q and 2q coulombs are located on the x-axis at x = 0.5a and 1.5a respectively, as shown above. There is an infinite grounded conducting plane at x = 0. What is the magnitude of the net force on the charge q?

A. 

B. 

C. 

D. 

E. 

(GR9277 #10)

Solution:

Using method of image:

On q:



with  

Answer: E

Electromagnetism - Electric Potential

Two large conducting plates form a wedge of angle α as shown-in the diagram above. The plates are insulated from each other; one has a potential V₀ and the other is grounded. Assuming that the plates are large enough so that the potential difference between them is independent of the cylindrical coordinates z and ρ, the potential anywhere between the plates as s function of the angle φ is

A. V₀/α
B. V₀φ/α
C. V₀α/φ
D. V₀φ²/α
E. V₀α/φ²

(GR9277 #12)

Solution:

Boundary conditions:
V(φ = 0) = 0 
V(φ = α) = V₀ 

(A) FALSE
V₀/α does not depend on φ 

(B) TRUE
V₀φ/α = 0 for φ = 0
V₀φ/α = V₀ for φ = α

(C) FALSE. V₀α/φ = ∞ for φ = 0

(D) FALSE
For φ = α, V₀φ²/α  = V₀α 

(E) FALSE
For φ = α, V₀φ²/α  = V₀α

Answer: B

Electromagnetism - Conductor

A conducting cavity is driven as an electromagnetic resonator. If perfect conductivity is assumed, the transverse and normal field components must obey which of the following conditions at the inner cavity walls?

A. E_n = 0, B_n = 0
B. E_n = 0, B_t = 0
C. E_t = 0, B_t = 0
D. E_t = 0, B_n = 0
E. None of the above

(GR9277 #34)

Solution:

Boundary Condition on the surface of perfect conductor:

• E_tangential = 0
• E_normal ≠ 0 → (A) and (B) FALSE
• B_tangential ≠ 0 → (C) FALSE
• B_normal = 0

Answer: D

Electromagnetism - Conductor

A plane-polarized electromagnetic wave is incident normally on a flat, perfectly conducting surface. Upon reflection at the surface, which of the following is true? 

A. Both the electric vector and magnetic vector are reversed.
B. Neither the electric vector nor the magnetic vector is reversed.
C. The electric vector is reversed; the magnetic vector is not.
D. The magnetic vector is reversed; the electric vector is not.
E.  The directions of the electric and magnetic vectors are interchanged.

(GR9277 #36)

Solution:

Perfect conductor: 

Its charge and current are distributed on the surface so that external fields can not penetrate it.

Boundary Condition (BC) on the surface of perfect conductor:

• E_tangential = 0
• E_normal ≠ 0 
• B_tangential ≠ 0
• B_normal = 0 

EM wave: E ⊥ B (perpendicular to each other).

If E normal to the surface of conductor, B is tangential.
BC: E_normal ≠ 0 and B_tangential ≠ 0
Only E is reversed.
B is not reserved, it's parallel to the surface.

Answer: C

Condensed Matter - Conductor and Semiconductor

Which of the following statements concerning the electrical conductivities at a room temperature of a pure copper sample and a pure silicon sample is NOT true? 

1. The conductivity of the copper sample is many orders of magnitude greater than that of silicon sample.
2. If the temperature of the copper sample is increased, its conductivity will decrease.
3. If the temperature of the silicon sample is increased, its conductivity will increase.
4. The addition of an impurity in the copper sample always decreases its conductivity.
5. The addition of an impurity in the silicon sample always decreases its conductivity.

(GR8677 #23)

Solution:

A. TRUE.
Copper → good conductor
Silicon → semi conductor

B. TRUE.
Conductivity is the inverse of resistivity, 
Copper → good conductor → resistivity, → , decreases.

C. TRUE.
Silicon → semi conductor → resistivity, → , exponentially increases

D. TRUE.
For conductor, doping will decrease conductivity.

E. FALSE.
For semiconductor, doping increases the charge carrier density (electron or electron holes) in the conduction band → increasing conductivity

Answer: E

Electromagnetism - Dielectric

A dielectric of dielectric constant K is placed in contact with a conductor having surface charge density σ, as shown above. What is the polarization (bound) charge density σ_p on the surface of the dielectric at the interface between the two materials? 

A. σK / (1−K)
B. K / (1 + K)
C. σK
D. σ(1 + K) / K
E. σ(1 − K) / K

(GR8677 #54)

Solution:

Polarization, P = ɛ₀χ_eE = ɛ₀(K − 1)E
Thus, K  1.
If K = 1 → P = 0 and σ_p = 0 → no polarized dielectric

Check answers for K = 1→ σ_p = 0

(A) σ_p = σK/(1 − K) = ∞ → FALSE
(B) σ_p = K/(1 + K) = 1/2 → FALSE
(C) σ_p = σK = σ → FALSE
(D) σ_p = σ(1 + K)/K = 2σ → FALSE
(E) σ_p = σ(1 − K)/K = 0 → TRUE

Answer: E 


Alternative Solution:

In dielectric, E = σ/ɛ = σ/ɛ₀K
Polarization, P = ɛ₀(K − 1)E
σ_p = ɛ₀(K − 1)(σ/ɛ₀K)
= (K − 1)(σ/K)
= σ(1 − K)/K

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 - Conductor

An electromagnetic plane wave, propagating in vacuum, has an electric field given by E = E₀ cos (kx−ωt) and is normally incident on a perfect conductor at x = 0, as shown in figure. Immediately to the left of the conductor, the total electric field E and the total magnetic field B are given by which of the following?

	E	B
A.	0	0
B.	2E0 cos ωt	0
C.	0	2(E0/c) cos ωt
D.	2E0 cos ωt	2(E0/c) cos ωt
E.	2E0 cos ωt	2(E0/c) sin ωt

(GR0177 #61)

Solution:

Perfect conductor: 

Its charge and current are distributed on the surface so that external fields can not penetrate it.

Boundary Condition (BC) on the surface of perfect conductor:

• E_tangential = 0
• E_normal ≠ 0 
• B_tangential ≠ 0
• B_normal = 0 

EM wave: E ⊥ B (perpendicular to each other).

If E normal to the surface of conductor, B is tangential.

BC: E_normal ≠ 0

E_initial is reflected by the conductor with the same magnitude but opposite direction.

Total E = E_initial − E_reflected = 0

(B), (D), (E) FALSE

BC: B_tangential ≠ 0

B is not reflected, it's parallel to the surface.

Total B ≠ 0

(A) FALSE

Answer: C