# SSC JE Electrical Previous Year Question Paper 2018-SET2|MES Electrical

Ques.21. The SI unit of the Magnetic flux is___

1. Henry
2. Coulomb
3. Tesla
4. Weber

S.I. unit of magnetic flux is weber abbreviated as Wb and C.G.S. unit of magnetic flux is maxwell abbreviated as Mx. These are defined as follows:

When the magnetic field of one Tesla passes normally through 1 m2 area of a surface then magnetic flux linked with this surface is defined as 1 Weber.

Ques.22. The relative permeability of the diamagnetic material is_____

1. Greater than 1
2. Greater than 10
3. Less than 1
4. Greater than 100

Permeability is a measure of how easy it is to establish the flux in a material. Ferromagnetic materials have high permeability and hence low Reluctance, while nonmagnetic materials have low permeability and high Reluctance.

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.

An interesting aspect of diamagnetism is the fact that the magnetic flux density inside the diamagnetic material is lower than the external magnetic field. If we place a piece of diamagnetic material over a permanent magnet, the magnet will repel the diamagnetic material, as shown in Figure.

Since the magnet and the equivalent magnetic field (due to the magnetization of the diamagnet material) oppose each other, the diamagnetic material is always repelled from the magnetic field in the same way that two magnets repel each other when their magnetic flux densities oppose each other. However, this force is extremely small for diamagnetic materials, except superconductors, in which it is very large. This repulsion is the reason why a permanent magnet floats above a superconducting material.

Ques.23. Which of the following is the reciprocal of the reluctance?

1. Permeability
2. Susceptibility
3. Permeance
4. Reluctivity

Permeance

The reciprocal of reluctance is called permeance and is a measure of the readiness with which the magnetic flux is developed. it is analogous to conductance in an electric circuit and is measured by Weber/ampere-turn.

Permeance = 1/Reluctance

∴ Permeance = flux ⁄ m.m.f

Ques.24. Which of the following is the CORRECT formula for permeance?

1. 1/H
2. B/H
3. φ/Fm
4. Fm

### Reluctance

Similar to the resistance characteristic in electricity, we have reluctance in magnetism which is the property of the substance that opposes the magnetic flux through it.

The reluctance of any part of a magnetic circuit may be defined as the ratio of the drop in magnetomotive force to the flux produced in that part of the circuit. It is measured in ampere-turns/Weber and is denoted by S.

Reluctance = m.m.f ⁄ flux

Reluctance is dependent on:

1. Nature of the substance.
2. Area of the cross-section of the material through which flux is passing (a).
3. Length of the magnetic path.

### Permeance

The reciprocal of reluctance is called permeance and is a measure of the readiness with which the magnetic flux is developed. it is analogous to conductance in an electric circuit and is measured by Weber/ampere-turn.

Permeance = 1/Reluctance

∴ Permeance = flux ⁄ m.m.f

Ques.25. Determine the intensity of the magnetization (in A/m) of a magnet when it has a pole strength of 60 A-m and a pole area of 20 sq.m

1. 9
2. 4
3. 6
4. 3

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 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 = 60 A-m

Pole area = 20 m2

Intensity = 60 ⁄ 20 = 3 A/m

Ques.26. What will be the magnitude of induced EMF (in V) in a coil that has 200 square loops, each of the side 5 cm and placed normal to a magnetic field? The magnetic field increases at the rate of 4 Weber per sq. meter.

1. 1
2. 2
3. 4
4. 3

If N is the number of turns in the coil, f is the frequency of rotation, A is the area of coil and B is the magnetic induction then induced emf is given by

e = −dφ/dt

Flux linkage through the coil is

φ = NBA cosωt

or

e = −dNBA cosωt ⁄ dφ

So According to the Faraday’s Law whenever there is a change in the magnetic flux linked with a coil, a] instantaneous emf is induced in the coil. The magnitude of the emf induced in the coil is proportional to the rate of variation of the magnetic flux linked with the coil:

e = −dNBA cosωt ⁄ dφ

Number of turns N = 200

Area A = 0.052

The rate of change of flux dφ ⁄ dt = 4

∴ Magnitude of the induced E.M.F is

E = 200 × 0.052 × 4

E = 2V

Ques.27. What will be the value of the current (in Ampere) in a 40 cm long solenoid in a free space, if it has 400 turns, 2 cm of diameter and produces a magnetic field of 4mT at its center?

1. 4.23
2. 5.15
3. 3.18
4. 2.34

The magnetic flux density at the center of a solenoid is given as

$B = \frac{{{\mu _o}NI}}{L}$

Where

B = Magnetic flux density = 4 × 10−3

N = Numbers of turns = 400 turn

I = Current = ?

L = Length of the solenoid = 40 cm = 0.4 m

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

$\begin{array}{l}4 \times {10^{ - 3}} = 4\pi \times {10^{ - 7}} \times \dfrac{{400 \times I}}{{0.4}}\\\\I = 3.18A\end{array}$

Ques.28. What will be the self-inductance (in Henry) of a 2 m long air-core solenoid, if the diameter of the solenoid is 25 cm and has 600 turns?

1. 0.011
2. 0.045
3. 0.132
4. 0.645

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

$L = \dfrac{{{\mu _o}{N^2}A}}{L}$

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

Length L = 2 m

Number of turns N = 600

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

$\begin{array}{l}L = \dfrac{{{{600}^2} \times \pi \times {{0.25}^2} \times 4\pi \times {{10}^{ - 7}}}}{{4 \times 2}}\\\\L = \dfrac{{36 \times {{10}^{ - 3}} \times {{0.25}^2} \times 4{\pi ^2}}}{8}\\\\L = 0.011H\end{array}$

Ques.29. What will be the mutual inductance (in mH) between the two coils?  If a current of 2 sin (100 πt) passes through one of the coils, which induces a maximum EMF of 10 πV in the second coil?

1. 40
2. 20
3. 50
4. 60

The voltage across an inductor is

V = Ldi/dt

where V = 10π

I = 2 sin (100πt)

Comparing it with the standard equation of sinusoidal current i.e I = Im sin ωt

I = 2 A

ω = 100π

$\begin{array}{l}V = L\dfrac{{di}}{{dt}}\\\\V = L\dfrac{d}{{dt}}{I_m}\sin \omega t\\\\V = L \times {I_m} \times \omega \times \cos \omega t\end{array}$

The voltage will be maximum when cosωt = 1

Vmax = L × Im × ω

10π = L × 2 × 100π

L = 0.05 H

or

L = 50 mH

Ques.30. Determine the magnitude of the EMF (in V) induced between the axis of rotation and the rim of the disc, when the disc of the diameter 40 cm rotates with an angular velocity of 40 revolutions per second and placed in a magnetic field of 1 T acting parallel to the rotation of the disc.

1. 6
2. 3.6
3. 4.8
4. 5.02

Speed in the rotating disk

ξ = ½ ω.B.R2

ω = angular velocity = 2πN = 2 × 3.14 × 40 = 251.2 rad/sec

B = Magnetic Field = 1 T

Radius R = 40/2 cm = 0.2 m

Applying the above formula

ξ = ½ 251.2 × 1 × 0.22

ξ = 5.02 V

### 4 thoughts on “SSC JE Electrical Previous Year Question Paper 2018-SET2|MES Electrical”

1. sir ek request hai kripya aap answer ko hide kar dijiye taki as a practise ise solve iske baad click karne pe answer dikhaye. ye achha huga practise ke liye.