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

Answer.3. Lead

Explanation:-

The magnetic materials can be classified on the basis of magnetic property namely relative permeability μ_r. In general, a material is said to be non-magnetic if the value of μ_r is less than 1 or approximately equal to 1. While the material is said to be magnetic if the value of μ_r 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 μ_r is slightly less than unity then it is a diamagnetic material (μ_r < 1).

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

Diamagnetic Material

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.

Example of diamagnetic materials are:-

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

Note:-

Oxygen is Paramagnetic

Aluminum is paramagnetic

Nickel is Ferromagnetic

Answer.3. Titanium

Explanation:-

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.

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

Note:-

Iron is ferromagnetic

Copper is Diamagnetic

Silver is Diamagnetic

Answer.1. φE = q/ε_o

Explanation:-

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/4πε_or² 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

φ_E = E × Area = (Q × 4πr²) ⁄ 4πε_or²

φ_E = Q/ε_o

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/ε_o times the charge enclosed.

Answer.4. 1.43

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 = 100 A-m

Pole area = 70 m²

Intensity = 100 ⁄ 70 = 1.43 A/m

Answer.4. Reluctivity

Explanation:-

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.

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 = K_h × B_M^1.67 × f × v watts

where
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.

Answer.2. 4

Explanation:-

Eddy current losses in the transformer is given by

Eddy current losses = K_e × B_m² × f² × V ×t²

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

Now put the above data in the equation we get

E_Loss= 1 × 2² × (50)² × 0.02² × 1

E_Loss = 4

B_m = 4 watt

Answer.1. 5.54

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 = 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]

Answer.3. 1

Explanation:-

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

K = M/√L₁L₂

where

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

Answer.4. 1.25

Explanation:-

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