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

Ques.1. Determine the energy (in J) stored by a 0.4 H inductance, If the current flowing through it is 2A. (SSC-2018, Set-1)

1. 1.6
2. 0.8
3. 0.4
4. 1.4

Explanation:-

The energy in a Magnetic Field of inductance

The magnetic flux of the current in inductance is the electric energy supplied by the voltage source producing the current. The energy is stored in the magnetic field since it can do the work of producing the induced voltage when the flux moves. The amount of electric energy stored is

Energy = ½LI2 joule

Given

Inductance H = 0.4
Current I = 2A

E = ½ × 0.4 × 22

E = 0.8 Joule

Ques.2. What will be the voltage (in V) across an 8H inductor, when the rate of change of current flowing through it is 0.5A/sec. (SSC-2018, Set-1)

1. 2
2. 6
3. 4
4. 8

Explanation:-

Let “I” be the current through the circuit and (di/dt) be the rate of change of current through the circuit at that instant.

∴ The instantaneous voltage across Inductor L
V = L(di/dt)

Given Inductance = 8H
Rate of change of current di/dt = 0.5A

V = 8 × 0.5 = 4 V

Instantaneous voltage across Inductor V = 4 V

Ques.3. The relative permeability of the diamagnetic material is (SSC-2018, Set-1)

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

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 (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.4. Which of the following is the CORRECT formula for permeance? (SSC-2018, Set-1,2)

1. I/H
2. B/H
3. φ/Fm
4. Fm/L

Explanation:-

### 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.5. Which one of the following is the S.I unit of the magnetic field strength? (SSC-2018, Set-1)

1. Weber
2. Tesla
3. Ampere-Meter
4. Ampere/Meter

Explanation:-

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.

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 the 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 symbol H and is measured in ampere-turns per meter (At/m).

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

Hence the magnetic field Intensity can be defined as the ratio of applied MMF to the length of the path that it acts over.

Note:- Magnetic ﬁeld intensity, H, is independent of the medium. Its value depends only on the number of turns N and the current I ﬂowing in the coil.

Magnetic Field Strength is equivalent to the Voltage gradient in an Electrical Circuit.

Ques.6. If the magnetic susceptibility of any material is less than zero then the material is—— (SSC-2018, Set-1)

1. Paramagnetic
2. Ferromagnetic
3. Diamagnetic
4. Ferrimagnetic

Explanation:-

The magnetic susceptibility expresses the responsiveness of a material to the applied magnetic field and can vary with the external applied magnetic field.

The symbol for magnetic susceptibility is χ and magnetic susceptibility is a number with no units.

Materials with positive susceptibility have a field-induced within them that results in a total field that is greater than the strength of the magnetic field in which they are placed. Such materials are described as paramagnetic.

Materials with negative susceptibility have an induced field the opposes the field in which the material has been placed, so the total field within the material is lower than the field in which they are placed. Such materials are described as diamagnetic.

For materials with positive susceptibility, the material is pushed toward stronger magnetic fields as it tends to lower potential energy whereas an opposite response for materials with negative susceptibility. Hence, materials with negative susceptibility (diamagnetic materials) are repelled and pushed toward weaker magnetic fields and materials with positive susceptibility (paramagnetic, ferromagnetic, ferromagnetic, and antiferromagnetic) are pulled toward stronger magnetic fields.

Ques.7. Determine the reluctance (in Amp-turns/Wb) of a coil, when the flux through the coil is 15 Wb and the produced MMF is 30 Amp-turns. (SSC-2018, Set-1)

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

Explanation:-

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 = 30 ⁄ 15

Reluctance = 2 Amp-turns/Wb

Ques.8. Determine the value of produced MMF (in Amp-turns) in a coil if the coil has 120 turns carrying a current of 0.1 A. (SSC-2018, Set-1)

1. 12
2. 14
3. 16
4. 18

Explanation:-

Magnetomotive force (MMF): It is defined as the force that drives magnetic flux throng the magnetic circuit. In all practical magnetic circuits, this is provided using a current-carrying winding (coil) and is the product of the current and the number of turns in the winding. Its SI unit of measurement is the ampere.

m.m.f Fm = N.I Ampere-turn

where N is the number of conductors (or turns) and I is the current in amperes. The unit of MMF is sometimes expressed as ‘ampere-turns’. However, since ‘turns’ have no dimensions, the Sl unit of MMF is the amp.

Given

Number of turns N = 120 turns

Current I = 0.1

Fm = 120 × 0.1

Fm = 12

Ques.9. Determine the magnitude of 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 60 revolutions per second and placed in the magnetic field of 3 T acting parallel to the rotation of the disc. (SSC-2018 Set-1,2)

1. 6.69
2. 4.64
3. 6.67
4. 5.65

Explanation:-

Speed in the rotating disk is given as

ξ = ½ ω.B.R2

ω = angular velocity = 2πN = 2 × 3.14 × 60 = 376.8 rad/sec

B = Magnetic Field = 3 T

Radius R = 10 cm = 0.1 m

Applying the above formula

ξ = ½ 376.8 × 3 × 0.12

ξ = 5.65 V

Ques.10. What will be the value of current (in A) in a 50 cm long air-core solenoid, if the value of the magnetic field at the center of the solenoid is 5 mT and the solenoid has 300 turns? (SSC-2018, Set-1)

1. 6.63
2. 5.63
3. 4.36
4. 8.25

Explanation:-

The magnetic flux density at the center of an air core solenoid is given as

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

Where

B = Magnetic flux density = 5 × 10−3

N = Numbers of turns = 300 turn

I = Current = ?

L = Length of the solenoid = 50 cm = 0.5 m

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

$\begin{array}{l}5 \times {10^{ - 3}} = 4\pi \times {10^{ - 7}} \times \dfrac{{300 \times I}}{{0.5}}\\\\I = 6.63A\end{array}$