Courses

# Test: Magnetization

## 10 Questions MCQ Test Electromagnetic Theory | Test: Magnetization

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:

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:

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.

QUESTION: 3

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

Solution:

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:

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:

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:

Explanation: The susceptibility is given by χm = μr-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:

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.

QUESTION: 8

Very small and positive susceptibility is found in

Solution:

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:

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.

QUESTION: 10

Identify the diamagnetic material.

Solution: