SSC JE Magnetic-Circuit Solved Questions (2018-2009)

Ques.51. Which of the following is a diamagnetic material? (SSC-2018, Set-6)

  1. Aluminum
  2. Oxygen
  3. Lead
  4. Nickel

Answer.3. Lead


The magnetic materials can be classified on the basis of magnetic property namely relative permeability μr. In general, a material is said to be non-magnetic if the value of μr is less than 1 or approximately equal to 1. While the material is said to be magnetic if the value of μr is greater than or equal to 1.

Generally, the magnetic materials are classified into three major parts namely diamagnetic, paramagnetic and ferromagnetic. If a value of μis slightly less than unity then it is a diamagnetic material (μ< 1).

If the value of μr is slightly greater than unity, then it is paramagnetic material (μr ≥ 1). If the value of μis very large than unity (μr>>1), then it is the ferromagnetic material.

Diamagnetic Material

Diamagnetic materials are materials with relative permeabilities slightly smaller than 1 (μr < 1). This class includes important materials such as mercury, gold, silver, copper, lead, silicon, and water. The relative permeability of most diamagnetic materials varies between 0.9999 and 0.99999 (susceptibility varies between (-10-5 and -104), and for most applications, they may be assumed to be nonmagnetic.

Example of diamagnetic materials are:-

Silver, lead, copper, water, mercury, carbon, etc.


Oxygen is Paramagnetic

Aluminum is paramagnetic

Nickel is Ferromagnetic


Ques.52. Which of the following material shows paramagnetism? (SSC-2018, Set-6)

  1. Copper
  2. Iron
  3. Titanium
  4. Silver

Answer.3. Titanium


The substances, which when placed in an external magnetizing field, get magnetized feebly in the direction of the magnetizing field are called paramagnetic. Paramagnetism is due to the presence of unpaired electrons in the material, so all atoms with incompletely filled atomic orbitals are paramagnetic. Due to their spin, unpaired electrons have a magnetic dipole moment and act like tiny magnets.

Paramagnetic material

Examples of paramagnetic substances are Aluminium, antimony, platinum, manganese, sodium, chromium, copper chloride, liquid oxygen, titanium etc.


Iron is ferromagnetic

Copper is Diamagnetic

Silver is Diamagnetic


Ques.54. Which of the following is the CORRECT expression for Gauss’ Law? (SSC-2018, Set-6)

  1. φE = q/εo
  2. φE = εo/q
  3. φE = 4E/H
  4. φE = 4H/E

Answer.1. φE = q/εo


Gauss Theorem

This theorem was first expressed by a German scientist Karl Fredrich Gauss (1777-1855) and maybe stated as under:

The electric flux passing through a closed surface surrounding a number of charges is equal to the algebraic sum of the charges inside the closed surface.

Consider a point charge Q lying at the center of a sphere of radius r which surrounds it completely as shown in the fig. The total number of tubes off flux originating from the charge is Q  and are normal to the surface of the sphere. The electric field E which equals Q/4πεor2 is also normal to the surface. As said earlier, the total number of lines of force passing perpendicularly through the whole surface of the sphere is

Gauss law

φE = E × Area = (Q × 4πr2) ⁄ 4πεor2

φE = Q/εo

Hence, Gauss’s law may be stated as under:

if a closed surface encloses a net charge (Q), then the surface integral of an electric field (E) over the closed surface is equal to 1/εo times the charge enclosed.


Ques.55. Determine the intensity of magnetization (in A/m) of a magnet, when its pole strength is 100 A-m and has a pole area of 70 sq. m. (SSC-2018, Set-6)

  1. 4.98
  2. 3.65
  3. 2.53
  4. 1.43

Answer.4. 1.43


The intensity of magnetization (I): The intensity of magnetization of a magnetized specimen is a measure of its ma~netisation. It is defined as the magnetic moment per unit volume, the specimen is so small that its magnetization can be supposed to be uniform. It is the degree to which a substance is magnetized when placed in a magnetic field It is denoted by I.

I = Magnetic Movement ⁄ Volume

If the specimen is of the uniform cross-sectional area (a), magnetic length 2L, and pole strength m,

then M = m × 2L and V = a × 2L.

Hence I = (m × 2L) ⁄ (a × 2L) = m ⁄ a

Thus, in the case of specimen Uniform cross-section, the intensity of magnetization is given by pole strength per unit face area.

I = Pole Strenght ⁄ Pole Area

Pole strength = 100 A-m

Pole area = 70 m2

Intensity = 100 ⁄ 70 = 1.43 A/m


Ques. 56. Hysteresis loss occurring in the material does NOT depend on which of the following parameters? (SSC-2018, Set-6)

  1. Hysteresis constant
  2. Magnetic flux density
  3. Frequency
  4. Reluctivity

Answer.4. Reluctivity


The word hysteresis has its origin in the Greek word hysterein which means “to lag behind.” Physically, hysteresis refers to a damping phenomenon in which two conjugate quantities such as stress and strain, magnetic induction and magnetizing field, etc. lag behind each other or become out of phase, thus damping some of the (mechanical. magnetic…) energy and thus heating the lattice. Mechanical and magnetic hystereses are the most well-known damping phenomena.

A transformer core is made up of ferromagnetic material and these materials are very sensitive to magnetization. These materials behave as magnates when the external magnetic field is applied. These materials having the number of domains in their structure and the domains are nothing but small permanent magnets whose axes are randomly oriented inside the material so net magnetization is zero.

Hysteresis Losses

But when an external magnetic field is applied, the axes of domains(small magnets) get aligned  to the axis of externally applied magnetic field and when this external magnetic field is removed, maximum domains attain their original position but some of them do not attain their position means material do not demagnetize completely and this is the reason for hysteresis loss. In the transformer, we give AC supply so after each half cycle reversal of the external magnetic field, so there is a reversal of domain, Hence extra work has to be done to completely reverse it and this extra work needs electrical energy which results in hysteresis loss.

Hysteresis Losses is given by

Hysteresis Loss = Kh × BM1.67 × f × v watts

Kh = Hysteresis constant depends upon the material
Bm = Maximum flux density
f = frequency
v = Volume of the core

Hence the Hysteresis Loss does not depend upon the “measure of the resistance of a material to the establishment of a magnetic field within it” i.e reluctivity.


Ques.57. Determine the eddy current loss (in W) in a material having eddy current coefficient of 1, the thickness of 0.02 m and a volume of 1 cubic meter, which is kept in a magnetic field of the maximum flux density of 2 T and supplied by a frequency of 50 Hz. (SSC-2018, Set-6)

  1. 3
  2. 4
  3. 5
  4. 6

Answer.2. 4


Eddy current losses in the transformer is given by

Eddy current losses = Ke × Bm2 × f2 × V ×t2

Ke = Eddy current constant = 1
t = thickness of the core = 0.02 m
V = Volume of material = 1 m3
Bm = Maximum flux density = 2 T 
f = frequency = 50 Hz
Eddy current losses = ?

Now put the above data in the equation we get

ELoss= 1 × 22 × (50)2 × 0.022 × 1

ELoss = 4

Bm = 4 watt


Ques.58. What will be the self-inductance (in mH) of a 4 m long air-core solenoid, if the diameter of the solenoid is 50 cm and has 300 turns? (SSC-2018, Set-6)

  1. 5.54
  2. 6.94
  3. 7.85
  4. 8.64

Answer.1. 5.54


The self-inductance of the air core solenoid is given as

[latex display=”true”]L = \dfrac{{{\mu _o}{N^2}A}}{L}[/latex]

µo = permeability of free space = 4π × 10-7

Length L = 4 m

Number of turns N = 300

In this case, we have an air core, therefore, it is round in the shape hence the area

A = πd2 ⁄ 4

Therefore Inductance L is

[latex]\begin{array}{l}L = \dfrac{{{{300}^2} \times \pi \times {{0.50}^2} \times 4\pi \times {{10}^{ – 7}}}}{{4 \times 4}}\\\\L = \dfrac{{9 \times {{10}^{ – 3}} \times {{0.50}^2} \times 4{\pi ^2}}}{16}\\\\L = 5.5mH\end{array}[/latex]


Ques.59. What is the coupling factor between the two coils, each having self-inductance of 25 mH and the mutual inductance between the two is 25 mH? (SSC-2018, Set-6)

  1. 0.5
  2. 0.75
  3. 1
  4. 2

Answer.3. 1


The coupling coefficient K is the degree or fraction of the magnetic coupling that occurs between the circuit.

K = M/√L1L2


K = coupling factor = ?

M = Mutual inductance = 25 mH

L1 = primary circuit self-Inductance = 25 mH

L2 = secondary circuit self-inductance = 25 mH

K = 25/√25 × 25

K = 1


Ques.60. Determine the magnitude of the EMF (in V) induced between the axis of rotation and the rim of the disc, when the disc of radius 10 cm rotates with an angular velocity of 10 revolutions per second, which is placed in a magnetic field of 4 T acting parallel to the rotation of the disc. (SSC-2018, Set-6)

  1. 0.5
  2. 0.75
  3. 1
  4. 1.25

Answer.4. 1.25


Speed in the rotating disk

ξ = ½ ω.B.R2

ω = angular velocity = 2πN = 2 × 3.14 × 10 = 62.8 rad/sec

B = Magnetic Field = 4 T

Radius R = 10 cm = 0.1 m

Applying the above formula

ξ = ½ 62.8 × 4 × 0.12

ξ = 1.25 V

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