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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Electromagnetic Fields Theory (EMFT)  >  Test: Magnetostatic Energy - Electrical Engineering (EE) MCQ

Test: Magnetostatic Energy - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Magnetostatic Energy

Test: Magnetostatic Energy for Electrical Engineering (EE) 2024 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Magnetostatic Energy questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Magnetostatic Energy MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Magnetostatic Energy below.
Solutions of Test: Magnetostatic Energy questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Magnetostatic Energy solutions in Hindi for Electromagnetic Fields Theory (EMFT) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Magnetostatic Energy | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Magnetostatic Energy - Question 1
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Find the induced EMF in an inductor of 2mH and the current rate is 2000 units.

Detailed Solution for Test: Magnetostatic Energy - Question 1

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

Test: Magnetostatic Energy - Question 2
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Find the work done in an inductor of 4H when a current 8A is passed through it?

Detailed Solution for Test: Magnetostatic Energy - Question 2

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.

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Test: Magnetostatic Energy - Question 3
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Find the inductance of a material with 100 turns, area 12 units and current of 2A in air.

Detailed Solution for Test: Magnetostatic Energy - Question 3

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

Test: Magnetostatic Energy - Question 4
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Calculate the magnetic energy when the magnetic intensity in air is given as 14.2 units(in 10-4 order)
Detailed Solution for Test: Magnetostatic Energy - Question 4

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.

Test: Magnetostatic Energy - Question 5
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Calculate the magnetic energy when the magnetic flux density is given by 32 units(in 108order)
Detailed Solution for Test: Magnetostatic Energy - Question 5

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.

Test: Magnetostatic Energy - Question 6
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Calculate the energy when the magnetic intensity and magnetic flux density are 15 and 65 respectively.
Detailed Solution for Test: Magnetostatic Energy - Question 6

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.

Test: Magnetostatic Energy - Question 7
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Find the inductance when the energy is given by 2 units with a current of 16A.
Detailed Solution for Test: Magnetostatic Energy - Question 7

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.

Test: Magnetostatic Energy - Question 8
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Find the power of an inductor of 5H and current 4.5A after 2 seconds.
Detailed Solution for Test: Magnetostatic Energy - Question 8

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.

Test: Magnetostatic Energy - Question 9
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Find the turns in an solenoid of inductance 23.4mH , current 2A and area 15cm.
Detailed Solution for Test: Magnetostatic Energy - Question 9

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.

Test: Magnetostatic Energy - Question 10
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The energy of a coil depends on the turns. State True/False. 
Detailed Solution for Test: Magnetostatic Energy - Question 10

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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