Electromagnetics practice questions

Answer: A) One-fourth as large

• ACorrect. Coulomb's law is inverse-square.
• BIncorrect. It's inverse-square.
• CIncorrect. Force decreases with distance.
• DIncorrect. It's inverse-square.

Answer: D) 899 V/m

• AIncorrect. Recheck the powers of ten.
• BIncorrect. Square the distance.
• CIncorrect. Recheck the powers of ten.
• DCorrect. \(kq/r^2 = 8.99\times10^9(10^{-9})/0.01\).

Answer: A) \(B^2/(2\mu_0)\)

Magnetic energy density is w = ½BH. With H = B/μ0 in free space, w = B²/(2μ0); the electric-field counterpart is ½ε0E².

• ACorrect. Magnetic energy density is w = ½BH. In free space H = B/μ0, so w = B²/(2μ0), in J/m³.
• BIncorrect. This is the energy density of an electric field, not a magnetic field.
• CIncorrect. The units of μ0B do not work out to J/m³, and energy density has to depend on the square of the field.
• DIncorrect. B/μ0 is the magnetic field intensity H in free space, measured in A/m. It is not an energy density.

Answer: D) 0.90 N

• AIncorrect. Recheck the powers of ten.
• BIncorrect. Square the distance.
• CIncorrect. Recheck the powers of ten.
• DCorrect. \(kq_1q_2/r^2 = 8.99\times10^9(10^{-12})/0.01\).

Answer: A) Makes its own magnetic field oppose the change in flux through the loop

Lenz's law says the induced current produces a magnetic field that opposes the change in flux that caused it. This is the source of the negative sign in Faraday's law, emf = −dΦ/dt.

• ACorrect. The induced current sets up a field that opposes the change in flux linkage that caused it. This is why Faraday's law has a negative sign: emf = −dΦ/dt.
• BIncorrect. Lenz's law opposes the change in flux, not the flux itself. If the external flux is decreasing, the induced field points in the same direction as the external field to keep the flux from falling.
• CIncorrect. The induced emf is proportional to the rate of change of flux (Faraday's law), not to the flux itself. Lenz's law sets the direction, not the size.
• DIncorrect. The force on the induced current opposes the motion that causes it. If it aided the motion, energy would be created from nothing, which would break conservation of energy.

Answer: A) The net electric charge enclosed by the surface

Gauss's law says the net electric flux out of any closed surface is proportional to the net charge inside it: ∮E·dS = Q_enc/ε₀.

• ACorrect. Gauss's law states that the closed-surface integral of D·dS equals the enclosed free charge, so the electric flux equals Q_enc/ε₀ in free space.
• BIncorrect. Current is related to the magnetic field around a closed path (Ampère's law), or to the rate of change of charge (the continuity equation). It is not related to electric flux by Gauss's law.
• CIncorrect. A changing magnetic flux is related to the induced EMF around a closed path (Faraday's law). It does not set the electric flux through a closed surface.
• DIncorrect. Gauss's law involves the electric field and enclosed charge, not the electric potential on the surface.

Answer: A) 0.1 H

• ACorrect. \(L = N\Phi/I\).
• BIncorrect. That multiplies by I.
• CIncorrect. That omits the turns.
• DIncorrect. That isn't the calculation.

Answer: C) 50,000 V/m

Between parallel plates the field is uniform, so E = V/d. After converting 2 mm to 0.002 m, E = 100/0.002 = 50,000 V/m.

• AIncorrect. This divides 100 V by 2 without converting the gap from millimeters to meters (2 mm = 0.002 m).
• BIncorrect. This multiplies the voltage by the gap in millimeters. The field is voltage divided by separation, not their product.
• CCorrect. For a uniform field, E = V/d = 100 V / 0.002 m = 50,000 V/m (50 kV/m).
• DIncorrect. This multiplies 100 V by 0.002 m instead of dividing.

Answer: A) Attractive

Parallel currents in the same direction attract and opposite currents repel. The force per unit length is μ0I1I2/(2πd), acting perpendicular to the wires.

• ACorrect. Each wire sits in the circular magnetic field of the other. Applying F = IL × B with both currents in the same direction gives a force on each wire that points toward the other wire. The force per unit length is μ0I1I2/(2πd).
• BIncorrect. The wires repel only when the currents flow in opposite directions, which flips the sign of IL × B.
• CIncorrect. Each current produces a magnetic field at the other wire, so there is a nonzero force of magnitude μ0I1I2/(2πd) per unit length.
• DIncorrect. F = IL × B is perpendicular to the current, so the force acts along the line joining the wires, not along their length.

Answer: D) 3 mH

Mutual inductance is the coupling coefficient times the geometric mean of the self-inductances: M = k√(L₁L₂) = 0.5 × 6 mH = 3 mH.

• AIncorrect. This is k·L₁·L₂ = 0.5 × 4 × 9 with no square root. The mutual inductance uses the geometric mean of the self-inductances.
• BIncorrect. This is √(L₁L₂) = √36 = 6 mH, which would only be the mutual inductance with perfect coupling (k = 1). It leaves out k = 0.5.
• CIncorrect. This multiplies k by the arithmetic mean (4 + 9)/2 = 6.5 mH. The formula uses the geometric mean √(L₁L₂).
• DCorrect. M = k√(L₁L₂) = 0.5 × √(4 × 9) mH = 0.5 × 6 mH = 3 mH.