Description

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

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.

Solution:

Answer: b

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.

QUESTION: 2

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

Solution:

Answer: a

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^{-7}x 5/(2π x 100 x 10^{-9}) = 10 units.

QUESTION: 3

The torque expression of a current carrying conductor is

Solution:

Answer: c

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.

QUESTION: 4

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

Solution:

Answer: a

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.

QUESTION: 5

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

Solution:

Answer: b

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.

QUESTION: 6

nd the permeability of a medium whose susceptibility is 100.

Solution:

Answer: d

Explanation: The susceptibility is given by χ_{m} = μ_{r}-1. To get permeability, μr = χm + 1 = 100 + 1 = 101 units.

QUESTION: 7

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

Solution:

Answer: a

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

QUESTION: 8

Very small and positive susceptibility is found in

Solution:

Answer: c

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

QUESTION: 9

Which of the following materials is ferrimagnetic?

Solution:

Answer: c

Explanation: Fe is iron and a ferromagnetic material. Sn and FeCl are not magnetic materials. The oxides of iron like ferric oxide Fe_{2}O_{3} is said to be a ferrimagnetic material.

QUESTION: 10

Identify the diamagnetic material.

Solution:

Answer: b

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.

### Magnetization and Magnetic Intensity

Video | 01:24 min

### L17 : Magnetization - Matter & Magnetism, Physics, Class 12

Video | 05:56 min

### Test

Doc | 1 Page

### Test

Doc | 1 Page

- Test: Magnetization
Test | 10 questions | 10 min

- Test: Cloze Test- 1
Test | 15 questions | 15 min

- Test: Cloze Test- 5
Test | 5 questions | 10 min

- Test: Cloze Test- 4
Test | 5 questions | 10 min

- Divisibility Test - Practice Test
Test | 5 questions | 10 min