# Electromagnetic Theory Questions and Answers – Magnostatic Energy

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This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Magnetostatic Energy”.

1. Find the induced EMF in an inductor of 2mH and the current rate is 2000 units.
a) 4
b) -4
c) 1
d) -1

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.

2. Find the work done in an inductor of 4H when a current 8A is passed through it?
a) 256
b) 128
c) 64
d) 512

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.

3. Find the inductance of a material with 100 turns, area 12 units and current of 2A in air.
a) 0.75mH
b) 7.5mH
c) 75mH
d) 753mH

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.
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4. Calculate the magnetic energy when the magnetic intensity in air is given as 14.2 units(in 10-4 order)
a) 1.26
b) 2.61
c) 6.12
d) 1.62

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.

5. Calculate the magnetic energy when the magnetic flux density is given by 32 units(in 108 order)
a) 4.07
b) 7.4
c) 0.47
d) 7.04

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.

6. Calculate the energy when the magnetic intensity and magnetic flux density are 15 and 65 respectively.
a) 755
b) 487.5
c) 922
d) 645

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.

7. Find the inductance when the energy is given by 2 units with a current of 16A.
a) 15.6mH
b) 16.5mH
c) 16.8mH
d) 15.8mH

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.

8. Find the power of an inductor of 5H and current 4.5A after 2 seconds.
a) 25.31
b) 50.62
c) 102.4
d) 204.8

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.

9. Find the turns in an solenoid of inductance 23.4mH , current 2A and area 15cm.
a) 900
b) 400
c) 498
d) 658

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.

10. The energy of a coil depends on the turns. State True/False.
a) True
b) False

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.

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