SSC JE Electrical Previous Year Question Paper 2018-SET 3|MES Electrical

Ques.21. What will happen to the inductance, if the number of turns of the coil is decreased?

  1. Increase
  2. Decrease
  3. Remain Same
  4. None of these

The inductance of the is given as

L = N2 ⁄ Reluctance

i.e L ∝ N2

Since the inductance is directly proportional to the number of turns in the coil, So if the number of turns will decrease the inductance will also decrease

The self-inductance of the coil can also be written as

L = μN2A/I = L1

where

N is the number of turns of the solenoid

A is the area of each turn of the coil

l is the length of the solenoid

and μ is the permeability constant

The inductance of a coil can be reduced by one of the following methods:

  • By taking some turns off the coil.
  • By stretching out of the coil until it has a greater length.
  • By decreasing the area of the coil since less coil area offers more opposition to the formation of magnetic field flux for a given amount of field force.
  • The way the turns are arranged:- a short thick coil will have higher self-inductance than a long, thin one.
  • Presence of a magnetic circuit:- If the coil is wound on an iron core, the same current will set up a greater magnetic flux and the self-inductance will be higher.

 

Ques.22. If the area of the cross-section is reduced to half of its value, then what will happen to the flux density?

  1. Doubled
  2. Halved
  3. Remain same
  4. One-fourth

Magnetic flux is the amount of magnetic field (or the number of lines of force) produced by a magnetic source. The symbol for magnetic flux is φ. The unit of magnetic flux is the Weber,* Wb.

Magnetic flux density is the amount of flux passing through a defined area that is perpendicular to the direction of the flux.

or

The ‘closeness’ of a field’s lines of the magnetic field indicates the intensity of the field at any given point. This is termed the magnet flux density (symbol: B) of the field at that particular point. Just as with a permanent magnet, the flux density is always greatest in the areas nearest the poles of the coil as shown in the figure.

Magnetic flux density

From Figure, it can be seen that more Lines of flux pass through the shaded rectangle near poles, than through the rectangle of the identical cross-sectional area placed midway along with the coil. So can define flux density as the ‘ flux per unit is expressed as follows:

B = φ/A Wb/m2 or tesla

From the above equation, the magnetic flux density is inversely proportional to the area hence if the area is reduced to half then the magnetic flux density will become double.

B = φ/1/2 = B = 2φ

 

Ques.23. Calculate the value of flux leakage coefficient, if φg is the air gap flux and φa is the flux in the iron core.

  1. φga
  2. φag
  3. φ× φa
  4. φ+ φa

The flux which does not follow the desired path of magnetic material is called leakage flux. Generally, it is not possible to confine the entire flux to the iron path only, unlike an electric current which can be confined to a wire by providing proper insulation. In any magnetic circuit, it is impossible to avoid magnetic leakage since there is no magnetic insulator available in practice. For example, the air is a good insulator of electricity, though, it conducts magnetic flux. The flux in the air gap is known as useful flux because it can be utilized for practical purposes.

The figure shows an iron ring wound with a coil and a narrow air gap. The total flux produced by the coil does not pass through the air gap as some of it leaks through the air surrounding the ring. These flux lines are called leakage flux.

Therefore the part of the magnetic flux that has its path within the magnetic circuit is known as the useful flux or main flux and that taking other paths is called leakage flux. This phenomenon of wastage of some flux is called magnetic leakage. The Sum of the two parts is called the total flux produced.

leakage

The ratio of the total flux produced by the magnet to the main flux is called leakage coefficient or leakage factor.

or

The ratio of total flux (flux in the iron path) to the useful flux (flux in the air gap) is known as the leakage factor (λ).

Mathematically, the leakage factor can be written as

Leakage factor λ = φT  ⁄ φm = Total useful flux ⁄ Main flux = (φL + φm) ⁄ φm

Where

φ= Leakage flux

φ= Main flux

φ= Total flux

or

Leakage factor λ = (Flux in the Iron Path) / (Flux in the air Gap)

λ = φa  ⁄ φg

 

Ques.24. The graph shown below represents the B-H curve for_______

B H curve

  1. Glass
  2. Steel
  3. Iron
  4. Air

B-H Curve (or Magnetisation Curve)

The B-H curve (or magnetization curve) indicates the manner in which the flux density (B) varies with the magnetizing force (H) and the relationship is given as

B = μrμoH

Where

 μo is called the permeability of free space or magnetic space constant. Its value is 4π × 10-7 H/m,

μ= Relative permeability of medium  which is equal to 1 for all non-magnetic material 

(i) For non-magnetic materials. For non-magnetic materials (e.g. air, copper, wood, etc.), the relationship between B and H is

B = μoH

Hence the B-H curve of a non-magnetic material is a straight line passing through the origin as shown in Fig. because the curve never saturates no matter how great the flux density may be. and a large m.m.f. is required to produce a given flux in the non-magnetic material (e.g. air), also µo = 4π x 10-7  is a constant quantity, therefore, B ∝ H

Air B H

(ii) For magnetic materials. For magnetic materials (e.g., iron, steel etc.), the relation between B and H is given by:

B = μrμoH

Here μr is not constant but varies with flux density. Consequently, the B-H curve of a magnetic material is non-linear. Fig. shows the general shape of the B-H curve of a magnetic material. The non-linearity of the curve indicates that the relative permeability r = B/μoH) of a magnetic material is not constant but depends upon flux density.

Magnetic Material B H

 

Ques.25. A conductor of length 0.6 m is situated at the right angle to a uniform magnetic field of flux density 4 Wb/sq.m. The conductor is moving with a velocity of 50m/s. Calculate the emf induced (in V) in the conductor.

  1. 100
  2. 115
  3. 118
  4. 120

The emf induced in a straight conductor of length l moving with velocity v perpendicular to a magnetic field B is

E.M.F induced in the moving conductor

E = Blvsinθ

Where
E is the generated voltage =?
B is the magnetic flux density = 4 Wb/m2
l is the length of a conductor in the magnetic field = 0.6 m
v is the velocity of the conductor perpendicular to the magnetic lines = 50m/s

The value of sinθ is maximum at 90 degrees i.e When the coil is perpendicular to the field axis. i.e

E = Blv

E = 4 × 0.6 × 50 = 120 V

E = 120 Volts

 

Ques.26. A flux of 2mWb passes through a strip of length and breadth 10 cm and 20 cm respectively. It is placed at an angle of 30 degrees to the direction of uniforms magnetic field. What is the magnetic flux density (in T)?

  1. 0.1
  2. 0.115
  3. 0.9
  4. 1

Accordingly, the magnetic flux φ passing through a surface of area A with a normal at an angle θ to a magnetic field of flux density B is given by:

φ = B.A.Cosθ

ans.26

Where

φ = Magnetic flux = 2mWb = 2 × 10−3 Wb

A = Area = (Length × Breadth) = (10 × 20) = 200 cm2 = 200 × 10−4 m2

θ = Angle to the magnetic field = 30°

B = Magnetic flux density =?

2 × 10−3  = B × 200 × 10−4 × Cos30°

B = (2 × 10−3 )  ⁄ (200 × 10−4) × 0.866

B = 0.115 Tesla or Wb/m2

 

Ques.27. A 600-turns coil has an inductance of 0.8 H. if one-fifth of the turns are removed, then determine the new inductance (in H).

  1. 0.512
  2. 1.25
  3. 2
  4. 2.5

The inductance of a coil of wire is given by

inductance of coil

Where

 N = Number of turns in the coil

μ = permeability of the core material

A = Area of the coil

l = length of the coil

Let the Initial Inductance and turns of the coil are L1 & N1 and the New length of the coil be L2 & N2

If one-fifth of the turns are removed, then the remaining turns are

N2 =(1 − 1/5) = 4/5.N1

Since L ∝ N2

∴ L1 ⁄ L2 = (N1 ⁄ N2)2

0.8 ⁄ L2 = (N1 ⁄ 4/5 × N1)2

0.8 ⁄ L2 = (5 ⁄ 4)2

L2 = 0.8 × 16 ⁄ 25 = 0.512 H

L2 = 0.512 H

 

Ques.28. A magnetic circuit of reluctance 3000 A-turns/Wb is wounded by a wire of 400 turns. Determine the average E.M.F induced (in V) in the coil, if the current of 2 A is flowing in the coil is reversed in 15s.

  1. 14.22
  2. 16.28
  3. 17.96
  4. 19.62

The Relation Between the Inductance and the Reluctance is

L = N2/ℜ

Where

N = Number of turns

ℜ = Reluctance of the coil

L = (400)2 ⁄ 3000

L = 53.33 H

Hence the Average Induced EMF will be

E = L.Δi ⁄ t

Where 

is the inductance of the coil = 53.33 H

t = time = 15 sec

Δi is the change in current in A

Now change in current

Δi = 2 − (−2) = 4A (since the current is reversed)

Putting all the value in the equation

E = (53.33 × 4) ⁄ 15

E = 14.22 Volt

 

Ques.29. For the figure shown below, the permeability in the core is 4 × 10-4 Wb/A-turns-m. What is the value of flux density (in T) in the core?

img.16o

 

  1. 0.2
  2. 0.382
  3. 0.6
  4. 1

MAGNETIC FIELD INTENSITY

Magnetic field Intensity (H) is also called as Magnetic field StrengthMagnetic IntensityMagnetic fieldMagnetic Force and Magnetization Force.

Magnetic Field Strength (H)  gives the quantitative measure of the strongness or weakness of the magnetic field.

magnetic field circuit

Suppose that a current of I amperes flows through a coil of N turns wound on a toroid of length I meters. The mmf is the total current linked to the magnetic circuit i.e  IN ampere-turns. If the magnetic circuit is homogeneous and of a uniform cross-sectional area, the MMF per meter length of the magnetic circuit is termed as the magneticfield strength, magnetic field intensity, or magnetizing force. It is represented by the symbol H and is measured in ampere-turns per meter (At/m).

$H = \dfrac{{NI}}{l}{\text{ AT/m}}$

Given

Current I = 5 A

Number of turns N = 60 turns

Relative permeability μr = 4 × 10-4

The field strength inside the solenoid is

$H = \dfrac{{NI}}{l}{\text{ AT/m}}$

Where N = Number of turns

I = current

l = length

l = πD = π × (10 × 10-2) = 0.314 m

$\begin{array}{l}H = \dfrac{{60 \times 5}}{{0.314}}{\rm{ }}\\\\{\text{ = 955.4 AT/m}}\end{array}$

Relation Between Magnetic Flux Density and Field Intensity

At any point in a magnetic field, field strength or field intensity H is the force maintaining the magnetic flux and producing a particular value of flux density B at that point. Hence the field intensity H is the cause and the flux density B the effect. Thus the flux density can be assumed proportional to field intensity in the magnetic field i.e. in free space,

B = μoH Tesla

B = 4 × 10-4 × 9.55 × 102

B = 0.382 Tesla

 

Ques.30. Calculate the total inductance (in H) of the circuit given below.

ques.30o

  1. 5.33
  2. 8.77
  3. 10.5
  4. 11.9
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