Free Electrical Machines MCQs with Answers

• A. 2 x 1(T2 N/m)
• B. 2 x KT3 N/m
• C. 2 x 10″5 N/m
• D. 2x 1(T7 N/m)

Explanation: For two parallel conductors, force per metre is F/L = μ₀I₁I₂/(2πd). With both currents 1 A and spacing 1 m, this becomes 2 × 10⁻⁷ N/m.

• A. cobalt
• B. chromium
• C. nickel
• D. tungsten

Explanation: Permalloy is a high-permeability nickel-iron alloy, with nickel as its main constituent.

• A. slightly less than unity
• B. equal to unity
• C. slightly more than unity
• D. equal to that ferromagnetic mate rials

Explanation: Paramagnetic materials are weakly attracted by a magnetic field, so their relative permeability is slightly greater than unity.

• A. electromotive force
• B. magnetomotive force
• C. conductance
• D. permittivity

Explanation: The gilbert is the CGS unit of magnetomotive force, analogous to the ampere-turn in SI usage.

• A. ferromagnetic
• B. paramagnetic
• C. diamagnetic
• D. bipolar

Explanation: Diamagnetic substances oppose an applied magnetic field and have permeability lower than the permeability of free space, so μᵣ < 1.

• A. voltage
• B. current
• C. magnetic field
• D. direction of force on the conductor

Explanation: In Fleming’s left-hand rule, the forefinger points in the direction of the magnetic field, the middle finger indicates current, and the…

• A. non-magnetic materials
• B. ferro-magnetic materials
• C. paramagnetic materials
• D. ferri-magnetic materials

Explanation: Ferrites are ceramic magnetic materials whose magnetic moments are oppositely aligned but unequal, producing ferrimagnetism.

• A. reluctivity
• B. susceptibility
• C. permittivity
• D. conductance

Explanation: Reluctivity is the reciprocal of permeability, so it is written as 1/μ.

• A. weber
• B. lumens
• C. tesla
• D. none of the above

Explanation: Magnetic flux density is flux per unit area, B = Φ/A, and its SI unit is the tesla.

• A. meter is surrounded by strong magnetic fields
• B. a soft iron shielding is used
• C. a plastic shielding is provided
• D. a shielding of anon-magnetic material is used

Explanation: Soft iron provides a low-reluctance path that diverts external stray magnetic flux around the meter mechanism.

• A. ferromagnetic materials
• B. ferrites
• C. non-ferrous materials
• D. diamagnetic materials

Explanation: Diamagnetic materials have relative permeability slightly below unity because they weakly oppose an applied magnetic field.

• A. Maxwell
• B. Telsa
• C. Weber
• D. All of the above

Explanation: A tesla measures magnetic flux density, since 1 T = 1 Wb/m². Maxwell and weber are units of magnetic flux, so tesla is the quantity that…

• A. 1 Wb/mm2
• B. 1 Wb/m
• C. 1 Wb/m2
• D. 1 mWb/m2

Explanation: By definition, magnetic flux density is measured in webers per square metre, so 1 tesla = 1 Wb/m².

• A. permeance
• B. residual magnetism
• C. susceptance
• D. reluctance

Explanation: When the magnetising field is removed, the remaining magnetism is called residual magnetism or remanence.

• A. Brass
• B. Copper
• C. Zinc
• D. Ebonite

Explanation: Ebonite is essentially non-magnetic, with permeability approximately equal to that of free space.

• A. non-magnetic substances
• B. diamagnetic substances
• C. ferromagnetic substances
• D. none of the above

Explanation: Magnetic susceptibility is positive for paramagnetic and ferromagnetic materials because they are attracted by an applied magnetic field.

• A. resistance of the coil
• B. motion of the magnet
• C. number of turns of the coil
• D. pole strength of the magnet

Explanation: By Faraday's law, induced emf depends on the rate of change of flux linkage, including motion, number of turns, and magnet strength.

• A. 240 N
• B. 24 N
• C. 2.4 N
• D. 0.24 N

Explanation: The force on a current-carrying conductor is F = BIL sinθ. With B = 0.8 T, I = 10 A, L = 0.3 m, and θ = 90°, F = 0.8 × 10 × 0.3 = 2.4 N.

• A. 2500 AT/m
• B. 250 AT/m
• C. 25 AT/m
• D. 2.5 AT/m

Explanation: Magnetic field strength is H = NI/l, so H = (50 × 1)/0.2 = 250 ampere-turns per metre.

• A. 100 N
• B. 10 N
• C. 1 N
• D. 0.1 N

Explanation: For two parallel conductors, force per metre is F/L = μ₀I₁I₂/(2πd). Substitution gives (4π × 10⁻⁷ × 100 × 100)/(2π × 0.02) = 0.1 N/m.