# Test: Magnetostatic Energy

## 10 Questions MCQ Test Electromagnetic Fields Theory | Test: Magnetostatic Energy

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Attempt Test: Magnetostatic Energy | 10 questions in 10 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
QUESTION: 1

### Find the induced EMF in an inductor of 2mH and the current rate is 2000 units.

Solution:

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

QUESTION: 2

### Find the work done in an inductor of 4H when a current 8A is passed through it?

Solution:

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.

QUESTION: 3

### Find the inductance of a material with 100 turns, area 12 units and current of 2A in air.

Solution:

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

QUESTION: 4

Calculate the magnetic energy when the magnetic intensity in air is given as 14.2 units(in 10-4 order)

Solution:

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.

QUESTION: 5

Calculate the magnetic energy when the magnetic flux density is given by 32 units(in 108order)

Solution:

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.

QUESTION: 6

Calculate the energy when the magnetic intensity and magnetic flux density are 15 and 65 respectively.

Solution:

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.

QUESTION: 7

Find the inductance when the energy is given by 2 units with a current of 16A.

Solution:

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.

QUESTION: 8

Find the power of an inductor of 5H and current 4.5A after 2 seconds.

Solution:

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.

QUESTION: 9

Find the turns in an solenoid of inductance 23.4mH , current 2A and area 15cm.

Solution:

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.

QUESTION: 10

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

Solution: