Test: Magnetostatic Energy - Electrical Engineering (EE) MCQ

Answer: b
Explanation: The induced emf is given by e = -Ldi/dt. Put L = 2 x 10^-3 and di/dt = 2000 in the equation. We get e = -2 x 10^-3 x 2000 = -4 units

Answer: b
Explanation: The work done in the inductor will be W = 0.5 x LI². On substituting L = 4 and I = 8, we get, W = 0.5 x 4 x 8² = 128 units.

Answer: a
Explanation: The inductance of any material(coil) is given by L = μ N²A/I. On substituting N = 100, A = 0.12 and I = 2, we get L = 4π x 10^-7 x 100² x 0.12/2 = 0.75 units

Answer: a
Explanation: The magnetic energy is given by E = 0.5 μ H². Put H = 14.2 and in air μ = 4π x 10^-7, we get E = 0.5 x 4π x 10^-7 x 14.2² = 1.26 x 10^-4 units.

Answer: a
Explanation: The magnetic energy is given by E = 0.5 μ H² and we know that μH = B. On substituting we get a formula E = 0.5 B²/μ. Put B = 32 and in air μ = 4π x 10^-7, we get E = 0.5 x 32²/4π x 10^-7 = 4.07 x 10⁸ units.

Answer: b
Explanation: The magnetic energy can also be written as E = 0.5 μH² = 0.5 BH, since B = μH. On substituting B = 65 and H = 15 we get E = 0.5 x 65 x 15 = 487.5 units.

Answer: a
Explanation: The energy stored in an inductor is given by E = 0.5 LI². To get L, put E = 2 and I = 16 and thus L = 2E/I² = 2 x 2/16² = 15.6mH.

Answer: a
Explanation: The energy stored in an inductor is given by E = 0.5 LI². Thus, put L = 5 and I = 4.5 and we get E = 0.5 x 5 x 4.5² = 50.625 units To get power P = E/t = 50.625/2 = 25.31 units.

Answer: c
Explanation: The inductance of any material(coil) is given by L = μ N²A/I.
Put L = 23.4 x 10^-3, I = 2 and A = 0.15, we get N as 498 turns.

Answer: a
Explanation: The inductance is directly proportional to square of the turns. Since the energy is directly proportional to the inductance, we can say both are dependent on each other.