Test: Magnetization - Electrical Engineering (EE) MCQ

# Test: Magnetization - Electrical Engineering (EE) MCQ

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## 10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Magnetization

Test: Magnetization for Electrical Engineering (EE) 2024 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Magnetization questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Magnetization MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Magnetization below.
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Test: Magnetization - Question 1

### Find the Lorentz force due to a conductor of length 2m carrying a current of 1.5A and magnetic flux density of 12 units.

Detailed Solution for Test: Magnetization - Question 1

Explanation: The Lorentz is given by the product of the current, differential length and the magnetic flux density. Put B = 12, I = 1.5 and L = 2, thus we get F = BIL = 12 x 1.5 x 2 = 36 units.

Test: Magnetization - Question 2

### Calculate the flux density due to a circular conductor of radius 100nm and current 5A in air.

Detailed Solution for Test: Magnetization - Question 2

Explanation: The field intensity of this conductor is I/2πR and since B = μH, the flux density will be B = μI/2πR. Put I = 5 and R = 100 x 10-9, thus we get B = 4π x 10-7x 5/(2π x 100 x 10-9) = 10 units.

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Test: Magnetization - Question 3

### The torque expression of a current carrying conductor is

Detailed Solution for Test: Magnetization - Question 3

Explanation: The torque is given by the product of the flux density, magnetic moment IA and the sine angle of the conductor held by the field. This gives T = BIA sin θ.

Test: Magnetization - Question 4

Find the current in a dipole with a moment of 16 units and area of 9 units.

Detailed Solution for Test: Magnetization - Question 4

Explanation: The dipole moment is given by M = IA. To get I, put M = 16 and A = 9, we get I = M/A = 16/9 = 1.78 units.

Test: Magnetization - Question 5

The expression for magnetization is given by(I-current, A-area, V-volume)

Detailed Solution for Test: Magnetization - Question 5

Explanation: The magnetization is defined as the magnetic moment per unit volume and the magnetic moment is IA. Thus M = IA/V is the expression.

Test: Magnetization - Question 6

nd the permeability of a medium whose susceptibility is 100.

Detailed Solution for Test: Magnetization - Question 6

Explanation: The susceptibility is given by χm = μr-1. To get permeability, μr = χm + 1 = 100 + 1 = 101 units.

Test: Magnetization - Question 7

Calculate the magnetization of a material with susceptibility of 50 and field intensity of 0.25 units.

Detailed Solution for Test: Magnetization - Question 7

Explanation: The magnetization is the product of the susceptibility and the field intensity given by M = χmH. Put χm = 50 and H = 0.25, then M = 50 x 0.25 = 12.5 units.

Test: Magnetization - Question 8

Very small and positive susceptibility is found in

Detailed Solution for Test: Magnetization - Question 8

Explanation: Paramagnetic materials are characterized by a small and positive susceptibility. The susceptibility and the temperature are directly related.

Test: Magnetization - Question 9

Which of the following materials is ferrimagnetic?

Detailed Solution for Test: Magnetization - Question 9

Explanation: Fe is iron and a ferromagnetic material. Sn and FeCl are not magnetic materials. The oxides of iron like ferric oxide Fe2O3 is said to be a ferrimagnetic material.

Test: Magnetization - Question 10

Identify the diamagnetic material.

Detailed Solution for Test: Magnetization - Question 10

Explanation: The diamagnetic materials are characterised by very small or negative susceptibility. Also the susceptibility is independent of the temperature. The material having these properties is germanium from the given options. Metals like gold and atoms with closed shells are also diamagnetic.

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