Test: Magnetostatic Energy


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


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This mock test of Test: Magnetostatic Energy for Electronics and Communication Engineering (ECE) helps you for every Electronics and Communication Engineering (ECE) entrance exam. This contains 10 Multiple Choice Questions for Electronics and Communication Engineering (ECE) Test: Magnetostatic Energy (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Magnetostatic Energy quiz give you a good mix of easy questions and tough questions. Electronics and Communication Engineering (ECE) students definitely take this Test: Magnetostatic Energy exercise for a better result in the exam. You can find other Test: Magnetostatic Energy extra questions, long questions & short questions for Electronics and Communication Engineering (ECE) on EduRev as well by searching above.
QUESTION: 1

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

Solution:

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

QUESTION: 2

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

Solution:

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.

QUESTION: 3

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

Solution:

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

QUESTION: 4

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

Solution:

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.

QUESTION: 5

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

Solution:

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.

QUESTION: 6

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

Solution:

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.

QUESTION: 7

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

Solution:

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.

QUESTION: 8

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

Solution:

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.

QUESTION: 9

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

Solution:

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.

QUESTION: 10

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

Solution:

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.

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