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

Answer.3. 1.94

Explanation:-

B_m = Maximum flux density = ? 
f = frequency = 50 Hz
Eddy current losses = 6W

Now put the above data in the equation we get

6 = 2 × B²m × (50)² × (4 × 10^-3)² × 20

B²_m = 3.75

B_m = 1.94

Answer.4. Weber

Explanation:-

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.

Answer.3. Less than 1

Explanation:-

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 non-magnetic materials have low permeability and high Reluctance.

Diamagnetic materials are materials with relative permeabilities slightly smaller than 1 (μ_r < 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 -10^–4), 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.

Permeance

Explanation:-

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

Answer.4. 3

Explanation:-

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 = 60 A-m

Pole area = 20 m²

Intensity = 60 ⁄ 20 = 3 A/m

Answer.B. 2

Explanation:-

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 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

∴ The magnitude of the induced E.M.F is

E = 200 × 0.05² × 4

E = 2V

Answer.3. 3.18

Explanation:-

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

[latex display=”true”]B = \frac{{{\mu _o}NI}}{L}[/latex]

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

µ_o = permeability of free space = 4π × 10^-7

[latex]\begin{array}{l}4 \times {10^{ – 3}} = 4\pi \times {10^{ – 7}} \times \dfrac{{400 \times I}}{{0.4}}\\\\I = 3.18A\end{array}[/latex]

Answer.1. 0.011

Explanation:-

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

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

µ_o = 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

[latex]\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}[/latex]

Answer.3. 50

Explanation:-

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

[latex]\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}[/latex]

The voltage will be maximum when cosωt = 1

∴ V_max = L × I_m × ω

10π = L × 2 × 100π

L = 0.05 H

or

L = 50 mH