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Electromagnetics & Magnetic Circuits (Electrical Engineering) Solved Questions & Notes (2026) - Apex Rankers

Engineering & Technology > Electrical Engineering > Electromagnetics & Magnetic Circuits

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Electromagnetics & Magnetic Circuits

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Q. 1 Electrical Engineering
Difficulty: Easy (1 Mark)
Faraday's Law of Electromagnetic Induction states that the magnitude of induced EMF in a circuit is directly proportional to:
A
The total magnetic flux linked with the circuit
B
The time rate of change of magnetic flux linkage (e = -N * dphi/dt)
✓ Correct
C
The cross-sectional area of the magnetic core
D
The permeability of free space
💡 Step-by-Step Explanation & Concept Rationale
Faraday's Law states that induced EMF e = -N * (dphi/dt). The magnitude depends strictly on how rapidly magnetic flux linkage changes with respect to time.
Q. 2 Electrical Engineering
Difficulty: Easy (1 Mark)
Lenz's Law, which determines the direction of an induced EMF and current, is an expression of:
A
Conservation of Momentum
B
Conservation of Energy
✓ Correct
C
Conservation of Charge
D
Gauss's Law
💡 Step-by-Step Explanation & Concept Rationale
Lenz's Law states that the induced current always flows in such a direction that its magnetic effect opposes the change in flux that produced it. This negative sign ensures that work must be done against electromagnetic forces to generate electrical energy, conforming to Conservation of Energy.
Q. 3 Electrical Engineering
Difficulty: Easy (1 Mark)
In a magnetic circuit, the opposition offered to the establishment of magnetic flux is called:
A
Permeance
B
Reluctance (S or R_m)
✓ Correct
C
Susceptance
D
Conductance
💡 Step-by-Step Explanation & Concept Rationale
Reluctance is the magnetic analogue of electric resistance. It is defined as S = l / (mu0 * mu_r * A), where l is the magnetic path length, A is cross-sectional area, and mu is permeability. Its unit is A-t/Wb or Henry^-1.
Q. 4 Electrical Engineering
Difficulty: Easy (1 Mark)
Permeance in a magnetic circuit is directly analogous to which parameter in an electric circuit?
A
Resistance
B
Conductance (G)
✓ Correct
C
Current
D
Capacitance
💡 Step-by-Step Explanation & Concept Rationale
Permeance is the reciprocal of reluctance (P = 1 / S = mu * A / l). It represents the ease with which magnetic flux is developed, analogous to electrical conductance G = 1 / R.
Q. 5 Electrical Engineering
Difficulty: Easy (1 Mark)
What is the SI unit of Magnetomotive Force (MMF)?
A
Tesla (T)
B
Ampere-turns (A-t) or Amperes (A)
✓ Correct
C
Weber (Wb)
D
Henry (H)
💡 Step-by-Step Explanation & Concept Rationale
MMF is the magnetic potential driving flux through a magnetic circuit: MMF = N * I (turns * current). Its SI unit is the Ampere-turn (A-t) or simply Ampere.
Q. 6 Electrical Engineering
Difficulty: Medium (1 Mark)
The energy stored per unit volume (energy density) in a magnetic field with flux density B and magnetic field intensity H is:
A
w = (1/2) * B * H = B^2 / (2 * mu)
✓ Correct
B
w = B * H
C
w = (1/2) * mu * B^2
D
w = H^2 / (2 * mu)
💡 Step-by-Step Explanation & Concept Rationale
Magnetic energy density is given by w = (1/2) * B * H. Since B = mu * H, this can be written as w = B^2 / (2 * mu) = (1/2) * mu * H^2 in Joules/m^3.
Q. 7 Electrical Engineering
Difficulty: Easy (1 Mark)
The energy stored in an inductor of inductance L carrying current I is given by:
A
W = (1/2) * L^2 * I
B
W = (1/2) * L * I^2
✓ Correct
C
W = L * I
D
W = (1/2) * (L / I)
💡 Step-by-Step Explanation & Concept Rationale
The work done in establishing current I through an inductor is W = integral(0 to I) L * i * di = (1/2) * L * I^2 Joules.
Q. 8 Electrical Engineering
Difficulty: Medium (1 Mark)
Hysteresis loss in a ferromagnetic material subjected to alternating magnetization is proportional to:
A
The square of the core thickness
B
The area enclosed by the B-H hysteresis loop
✓ Correct
C
The inverse of supply frequency
D
The square of the applied voltage
💡 Step-by-Step Explanation & Concept Rationale
The area enclosed by the B-H loop represents the energy lost as heat per unit volume per cycle of magnetization. According to Steinmetz's empirical formula: P_h = eta * B_max^1.6 * f * V.
Q. 9 Electrical Engineering
Difficulty: Medium (1 Mark)
Eddy current loss in a magnetic core subjected to an alternating magnetic field varies with frequency (f) and maximum flux density (Bm) as:
A
P_e proportional to f * Bm
B
P_e proportional to f^2 * Bm^2
✓ Correct
C
P_e proportional to f^2 * Bm
D
P_e proportional to f * Bm^2
💡 Step-by-Step Explanation & Concept Rationale
Eddy current loss is given by P_e = K_e * Bm^2 * f^2 * t^2 * V, where t is the thickness of laminations. It is directly proportional to the square of frequency and the square of maximum flux density.
Q. 10 Electrical Engineering
Difficulty: Easy (1 Mark)
Why are transformer and motor cores constructed using thin, insulated silicon steel laminations instead of solid iron blocks?
A
To eliminate hysteresis loss completely
B
To increase the mechanical rigidity of the core
C
To increase electrical resistance across the path of circulating currents, thereby minimizing eddy current loss
✓ Correct
D
To increase the saturation flux density Bm
💡 Step-by-Step Explanation & Concept Rationale
Laminating the core into thin sheets insulated by varnish restricts the path of circulating eddy currents to tiny loops within each lamination. Since eddy current loss is proportional to thickness squared (t^2), laminating dramatically reduces P_e.
Q. 11 Electrical Engineering
Difficulty: Medium (1 Mark)
Adding approximately 3% to 4% silicon to electrical core steel serves primarily to:
A
Increase electrical resistivity of the steel, thereby reducing eddy current losses and reducing hysteresis loss
✓ Correct
B
Make the core ductile and mechanically flexible
C
Eliminate the need for transformer oil cooling
D
Decrease magnetic permeability
💡 Step-by-Step Explanation & Concept Rationale
Silicon increases the electrical resistivity of the steel significantly, which suppresses eddy current flow. It also reduces magnetostriction and decreases the area of the hysteresis loop, lowering total core losses.
Q. 12 Electrical Engineering
Difficulty: Easy (1 Mark)
Residual magnetism (or remanence) on a B-H hysteresis curve corresponds to:
A
The magnetic field intensity H required to reduce B to zero
B
The magnetic flux density B remaining in the material when the magnetizing field H is reduced to zero
✓ Correct
C
The saturation flux density at infinite field strength
D
The permeability of the core at high temperatures
💡 Step-by-Step Explanation & Concept Rationale
Remanence (or retentivity) is the value of flux density B that persists in the ferromagnetic core after the external magnetizing field H has been removed (H = 0).
Q. 13 Electrical Engineering
Difficulty: Easy (1 Mark)
Coercive force (or coercivity) on a B-H curve represents:
A
The maximum flux density achievable in the core
B
The reverse magnetizing force (-H) required to demagnetize the material and reduce residual flux density to zero
✓ Correct
C
The magnetic flux leakage across the air gap
D
The initial slope of the magnetization curve
💡 Step-by-Step Explanation & Concept Rationale
Coercivity is the reverse magnetic field intensity required to completely wipe out the remanent magnetic flux density (bringing B back to 0). Permanent magnets require high coercivity, whereas transformer cores require low coercivity.
Q. 14 Electrical Engineering
Difficulty: Easy (1 Mark)
The force experienced by a straight current-carrying conductor of length L carrying current I placed in a uniform magnetic field B is given by:
A
F = I * (L x B) = B * I * L * sin(theta)
✓ Correct
B
F = B * I * L * cos(theta)
C
F = B^2 * I * L
D
F = (I / B) * L
💡 Step-by-Step Explanation & Concept Rationale
Lorentz magnetic force on a current element is dF = I * (dL x B). For a straight conductor, F = B * I * L * sin(theta), where theta is the angle between the conductor and the magnetic field vector. Force is maximum when theta = 90 degrees.
Q. 15 Electrical Engineering
Difficulty: Easy (1 Mark)
Fleming's Left-Hand Rule is used to determine the direction of:
A
Induced EMF in an AC generator
B
Force (motion) acting on a current-carrying conductor in a magnetic field (Electric Motor principle)
✓ Correct
C
Magnetic field around a straight wire
D
Eddy currents in a solid plate
💡 Step-by-Step Explanation & Concept Rationale
Fleming's Left-Hand Rule applies to motors: Thumb = Thrust/Force/Motion, Forefinger = Magnetic Field (B), Center finger = Current (I). Fleming's Right-Hand Rule applies to generators.
Q. 16 Electrical Engineering
Difficulty: Easy (1 Mark)
Fleming's Right-Hand Rule is used to determine the direction of:
A
Force on a motor armature
B
Dynamically induced EMF or current in a conductor moving across a magnetic field (Electric Generator principle)
✓ Correct
C
Torque in a three-phase induction motor
D
Magnetic polarity of a solenoid
💡 Step-by-Step Explanation & Concept Rationale
Fleming's Right-Hand Rule applies to generators: Thumb = Motion of conductor, Forefinger = Direction of Magnetic Field (N to S), Middle finger = Direction of Induced EMF/Current.
Q. 17 Electrical Engineering
Difficulty: Easy (1 Mark)
The dynamically induced EMF in a straight conductor of length L moving with velocity v at an angle theta to a uniform magnetic field B is:
A
e = B * L * v * cos(theta)
B
e = B * L * v * sin(theta)
✓ Correct
C
e = (1/2) * B * L * v^2
D
e = B^2 * L * v
💡 Step-by-Step Explanation & Concept Rationale
By Lorentz force on moving conduction electrons: e = integral (v x B) . dl = B * L * v * sin(theta). When the conductor moves perpendicular to the magnetic lines of force (theta = 90 deg), e_max = B * L * v.
Q. 18 Electrical Engineering
Difficulty: Medium (1 Mark)
The self-inductance L of a solenoid with N turns, core cross-sectional area A, core length l, and relative permeability mu_r is given by:
A
L = (mu0 * mu_r * N * A) / l
B
L = (mu0 * mu_r * N^2 * A) / l
✓ Correct
C
L = (mu0 * N^2 * l) / A
D
L = (N * A) / (mu0 * mu_r * l)
💡 Step-by-Step Explanation & Concept Rationale
L = N * phi / I. Since phi = B * A = (mu * N * I / l) * A, substituting gives L = (mu0 * mu_r * N^2 * A) / l = N^2 / S, proving that inductance is proportional to the square of turns (N^2).
Q. 19 Electrical Engineering
Difficulty: Easy (1 Mark)
If the number of turns in an inductor is doubled while maintaining the same physical core dimensions, its inductance becomes:
A
Doubled (2x)
B
Quadrupled (4x)
✓ Correct
C
Halved (0.5x)
D
Unchanged
💡 Step-by-Step Explanation & Concept Rationale
Because L is proportional to N^2 (L ∝ N^2), doubling the number of turns (N' = 2N) results in L' = (2N)^2 = 4 * N^2 = 4 * L.
Q. 20 Electrical Engineering
Difficulty: Easy (1 Mark)
The magnetic field intensity H inside an ideal infinitely long solenoid having n turns per unit length carrying current I is:
A
H = n * I
✓ Correct
B
H = n * I / (2 * pi)
C
H = mu0 * n * I
D
H = n^2 * I
💡 Step-by-Step Explanation & Concept Rationale
By Ampere's Circuital Law, the magnetic field intensity inside a long solenoid is H = n * I (in A/m). The magnetic flux density is B = mu * H = mu0 * mu_r * n * I.
Q. 21 Electrical Engineering
Difficulty: Medium (1 Mark)
Ampere's Circuital Law states that the line integral of magnetic field intensity H around any closed contour equals:
A
The total magnetic flux enclosed
B
The net electric current enclosed by the loop (∮ H . dl = I_enclosed)
✓ Correct
C
Zero under all conditions
D
The rate of change of electric flux
💡 Step-by-Step Explanation & Concept Rationale
Ampere's Circuital Law states that ∮ H . dl = I_enc. In Maxwell's generalized equation for time-varying fields, displacement current density is added: ∮ H . dl = I_conduction + d/dt(integral D . dA).
Q. 22 Electrical Engineering
Difficulty: Medium (1 Mark)
Gauss's Law for Magnetism (div B = 0 or ∮ B . dA = 0) mathematically implies that:
A
Magnetic field is conservative
B
Isolated magnetic monopoles do not exist in nature, and magnetic field lines form continuous closed loops
✓ Correct
C
Magnetic charge equals electric charge
D
Magnetic flux cannot penetrate ferromagnetic substances
💡 Step-by-Step Explanation & Concept Rationale
div B = 0 states that the divergence of magnetic flux density everywhere is zero. This means there are no sources or sinks of magnetic field; isolated magnetic monopoles (North or South alone) do not exist, and magnetic lines always close upon themselves.
Q. 23 Electrical Engineering
Difficulty: Easy (1 Mark)
The Biot-Savart Law allows calculating the magnetic field produced by a current element. The magnetic field dB at a distance r is proportional to:
A
r
B
1 / r
C
1 / r^2
✓ Correct
D
1 / r^3
💡 Step-by-Step Explanation & Concept Rationale
dB = (mu0 / (4*pi)) * (I * dl x r_hat) / r^2 = (mu0 / (4*pi)) * (I * dl * sin(theta)) / r^2. It follows an inverse-square law with respect to distance from the current element.
Q. 24 Electrical Engineering
Difficulty: Medium (1 Mark)
Two parallel conductors placed 1 meter apart in vacuum carry currents of 1 A in the same direction. What is the nature and magnitude of the force per unit length between them?
A
Repulsive force of 2 * 10^-7 N/m
B
Attractive force of 2 * 10^-7 N/m
✓ Correct
C
Attractive force of 4*pi * 10^-7 N/m
D
Zero force
💡 Step-by-Step Explanation & Concept Rationale
Parallel conductors carrying currents in the same direction attract each other with force per unit length F/L = (mu0 * I1 * I2) / (2 * pi * d) = (4*pi*10^-7 * 1 * 1) / (2 * pi * 1) = 2 * 10^-7 N/m. This formula officially defines the SI base unit Ampere.
Q. 25 Electrical Engineering
Difficulty: Easy (1 Mark)
Two parallel conductors carrying currents in opposite directions will:
A
Attract each other
B
Repel each other
✓ Correct
C
Experience zero magnetic interaction
D
Rotate until they are perpendicular
💡 Step-by-Step Explanation & Concept Rationale
When currents flow in opposite directions, the magnetic field between the conductors reinforces while the field outside weakens, creating a lateral pressure that repels the two conductors.
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