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.

A.
24
B.
36
C.
32
D.
45

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.

A.
10
B.
100
C.
0.1
D.
1

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^-7x 5/(2π x 100 x 10^-9) = 10 units.

QUESTION: 3

The torque expression of a current carrying conductor is

A.
T = BIA cos θ
B.
T = BA cos θ
C.
T = BIA sin θ
D.
T = BA sin θ

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.

A.
1.78
B.
2.78
C.
1.87
D.
2.34

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)

A.
M = IAV
B.
M = IA/V
C.
M = V/IA
D.
e)a) M = IAVb) M = IA/Vc) M = V/IA

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.

A.
-100
B.
99
C.
-99
D.
101

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.

A.
12.5
B.
25
C.
75
D.
37.5

Solution:

Answer: a
Explanation: The magnetization is the product of the susceptibility and the field intensity given by M = χ_mH. 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

A.
Ferromagnetic
B.
Diamagnetic
C.
Paramagnetic
D.
Antiferromagnetic

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?

A.
Fe
B.
Sn
C.
Fe₂O₃
D.
FeCl

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 Fe₂O₃ is said to be a ferrimagnetic material.

QUESTION: 10

Identify the diamagnetic material.

A.
Silicon
B.
Germanium
C.
Silver
D.
Cobalt

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.