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# Multiple choice question for engineering

## Set 1

1. The presence of parallel alignment of magnetic dipole moment is given by which materials?
a) Diamagnetic
b) Ferromagnetic
c) Paramagnetic
d) Ferromagnetic

Answer: b [Reason:] The ferromagnetic materials are characterized by parallel alignment of magnetic dipole moments. Their susceptibility is very large.

2. The magnetic materials follow which law?
b) Ampere law
c) Lenz law
d) Curie Weiss law

Answer: d [Reason:] Generally, the ferromagnetic, paramagnetic and diamagnetic materials follow the Curie Weiss law, which relates the magnetization and the applied field.

3. Find the internal field when the applied field is 12 units, molecular field constant is 0.1 units and the magnetization is 74 units.
a) 86
b) 62
c) 752
d) 19.4

Answer: d [Reason:] From Curie law, the internal field of a magnetic material is given by H = Ho + χ M, where χ is the molecular field constant. Put χ = 0.1, M = 74 and Ho = 12, we get H = 12 + (0.1)74 = 19.4 units.

4. In which materials the magnetic anisotropy is followed?
a) Diamagnetic
b) Paramagnetic
c) Ferromagnetic
d) Ferromagnetic

Answer: c [Reason:] In materials like iron, the magnetic properties depend on the direction in which they are measured. This is magnetic anisotropy. The material iron is a ferromagnetic material type.

5. Piezoelectric effect is analogous to which phenomenon?
a) Electrostriction
b) Magnetostriction
c) Anisotropy
d) Magnetization

Answer: b [Reason:] The piezoelectric effect is the mechanical strain caused on a material like quartz when subjected to an electric field. The same is observed in a ferromagnetic material called magnetostriction.

6. The converse of magnetostriction is called the
a) Magnetization
b) Magnetic anisotropy
c) Villari effect
d) Curie effect

Answer: c [Reason:] When a strain is applied, the change in magnetic field is observed. This is the converse of the magnetostriction phenomenon and is called Villari effect.

7. The materials having very small susceptibility at all temperatures are
a) Antiferromagnetic
b) Diamagnetic
c) Ferromagnetic
d) Paramagnetic

Answer: a [Reason:] In antiferromagnetic materials, the susceptibility will decrease with increase in temperature. They have relatively small susceptibility at all temperatures.

8. Find the susceptibility when the curie constant is 0.2 and the difference in critical temperature and paramagnetic curie temperature is 0.01.
a) 2
b) 20
c) 0.02
d) 200

Answer: b [Reason:] The susceptibility in magnetic materials is given by χm = C/(T-θ), where C is the curie constant, T is the critical temperature and θ is the paramagnetic curie temperature. Put C = 0.2 and T-θ = 0.01, thus we get susceptibility as 0.2/0.01 = 20.

9. The susceptibility is independent of temperature in which material?
a) Paramagnetic
b) Ferromagnetic
c) Diamagnetic
d) Ferromagnetic

Answer: c [Reason:] In the diamagnetic materials, the susceptibility is very small and negative. Thus the susceptibility will be independent of the temperature. The atoms of solids having closed shells and metals like gold have this property.

10. In ferromagnetic materials the susceptibility is infinity. State True/False
a) True
b) False

Answer: a [Reason:] The ferromagnetic materials are iron, nickel, cobalt which are highly attracted by magnetic field. Thus their susceptibility is also very high and nearing infinity. Also ferrimagnetics have infinite susceptibility.

## Set 2

1. The magnetic vector potential is a scalar quantity.
a) True
b) False

Answer: b [Reason:] The magnetic vector potential could be learnt as a scalar. But it is actually a vector quantity, which means it has both magnitude and direction.

2. Find the magnetic field intensity when the magnetic vector potential x i + 2y j + 3z k.
a) 6
b) -6
c) 0
d) 1

Answer: b [Reason:] The magnetic field intensity is given by H = -Grad(Vm). The gradient of Vm is 1 + 2 + 3 = 6. Thus H = -6 units.

3. The value of ∫ H.dL will be
a) J
b) I
c) B
d) H

Answer: b [Reason:] By Stoke’s theorem, ∫ H.dL = ∫ Curl(H).dS and from Ampere’s law, Curl(H) = J. Thus ∫ H.dL = ∫ J.dS which is nothing but current I.

4. Given the vector potential is 16 – 12sin y j. Find the field intensity at the origin.
a) 28
b) 16
c) 12
d) 4

Answer: c [Reason:] The field intensity is given by H = – Grad(V). The gradient is given by 0 – 12cos y. At the origin, the gradient will be -12 cos 0 = -12. Thus the field intensity will be 12 units.

5. Find the vector potential when the field intensity 60x2 varies from (0,0,0) to (1,0,0).
a) 120
b) -20
c) -180
d) 60

Answer: b [Reason:] The field intensity H = -Grad(V). To get V, integrate H with respect to the variable. Thus V = -∫H.dl = -∫60x2 dx = -20x3 as x = 0->1 to get -20.

6. Find the flux density B when the potential is given by x i + y j + z k in air.
a) 12π x 10-7
b) -12π x 10-7
c) 6π x 10-7
d) -6π x 10-7

Answer: b [Reason:] The field intensity H = -Grad(V). Since the given potential is a position vector, the gradient will be 3 and H = -3. Thus the flux density B = μH = 4π x 10-7 x (-3) = -12π x 10-7 units.

7. The Laplacian of the magnetic vector potential will be
a) –μ J
b) – μ I
c) –μ B
d) –μ H

Answer: a [Reason:] The Laplacian of the magnetic vector potential is given by Del2(A) = -μ J, where μ is the permeability and J is the current density.

8. The magnetic vector potential for a line current will be inversely proportional to
a) dL
b) I
c) J
d) R

Answer: d [Reason:] The magnetic vector potential for the line integral will be A = ∫ μIdL/4πR. It is clear that the potential is inversely proportional to the distance or radius R.

9. The current element of the magnetic vector potential for a surface current will be
a) J dS
b) I dL
c) K dS
d) J dV

Answer: c [Reason:] The magnetic vector potential for the surface integral is given by A = ∫ μKdS/4πR. It is clear that the current element is K dS.

10. The relation between flux density and vector potential is
a) B = Curl(A)
b) A = Curl(B)
c) B = Div(A)
d) A = Div(B)

Answer: a [Reason:] The magnetic flux density B can be expressed as the space derivative of the magnetic vector potential A. Thus B = Curl(A).

## Set 3

1. Find the Lorentz force due to a conductor of length 2m carrying a current of 1.5A and magnetic flux density of 12 units.
a) 24
b) 36
c) 32
d) 45

Answer: b [Reason:] The Lorentz is given by the product of the current, differential length and the magnetic flux density. Put B = 12, I = 1.5 and L = 2, thus we get F = BIL = 12 x 1.5 x 2 = 36 units.

2. Calculate the flux density due to a circular conductor of radius 100nm and current 5A in air.
a) 10
b) 100
c) 0.1
d) 1

Answer: a [Reason:] The field intensity of this conductor is I/2πR and since B = μH, the flux density will be B = μI/2πR. Put I = 5 and R = 100 x 10-9, thus we get B = 4π x 10-7x 5/(2π x 100 x 10-9) = 10 units.

3. The torque expression of a current carrying conductor is
a) T = BIA cos θ
b) T = BA cos θ
c) T = BIA sin θ
d) T = BA sin θ

Answer: c [Reason:] The torque is given by the product of the flux density, magnetic moment IA and the sine angle of the conductor held by the field. This gives T = BIA sin θ.

4. Find the current in a dipole with a moment of 16 units and area of 9 units.
a) 1.78
b) 2.78
c) 1.87
d) 2.34

Answer: a [Reason:] The dipole moment is given by M = IA. To get I, put M = 16 and A = 9, we get I = M/A = 16/9 = 1.78 units.

5. The expression for magnetization is given by(I-current, A-area, V-volume)
a) M = IAV
b) M = IA/V
c) M = V/IA
d) M = 1/IAV

Answer: b [Reason:] The magnetization is defined as the magnetic moment per unit volume and the magnetic moment is IA. Thus M = IA/V is the expression.

6. Find the permeability of a medium whose susceptibility is 100.
a) -100
b) 99
c) -99
d) 101

Answer: d [Reason:] The susceptibility is given by χm = μr-1. To get permeability, μr = χm + 1 = 100 + 1 = 101 units.

7. Calculate the magnetization of a material with susceptibility of 50 and field intensity of 0.25 units.
a) 12.5
b) 25
c) 75
d) 37.5

Answer: a [Reason:] The magnetization is the product of the susceptibility and the field intensity given by M = χmH. Put χm = 50 and H = 0.25, then M = 50 x 0.25 = 12.5 units.

8. Very small and positive susceptibility is found in
a) Ferromagnetic
b) Diamagnetic
c) Paramagnetic
d) Antiferromagnetic

Answer: c [Reason:] Paramagnetic materials are characterized by a small and positive susceptibility. The susceptibility and the temperature are directly related.

9. Which of the following materials is ferrimagnetic?
a) Fe
b) Sn
c) Fe2O3
d) FeCl

Answer: c [Reason:] Fe is iron and a ferromagnetic material. Sn and FeCl are not magnetic materials. The oxides of iron like ferric oxide Fe2O3 is said to be a ferrimagnetic material.

10. Identify the diamagnetic material.
a) Silicon
b) Germanium
c) Silver
d) Cobalt

Answer: b [Reason:] The diamagnetic materials are characterised by very small or negative susceptibility. Also the susceptibility is independent of the temperature. The material having these properties is germanium from the given options. Metals like gold and atoms with closed shells are also diamagnetic.

## Set 4

1. The magnetostatics highly relies on which property?
a) Resistance
b) Capacitance
c) Inductance
d) Moment

Answer: c [Reason:] The magnetostatics highly relies on the inductance of the magnetic materials, which decides its behavior in the influence of magnetic field.

2. The inductance is the measure of
a) Electric charges stored by the material
b) Emf generated by energising the coil
c) Magnetic field stored by the material
d) Magnetization of dipoles

Answer: b [Reason:] The inductance is a property of an electric conductor/coil which measures the amount of emf generated by passing current through the coil.

3. Find the total flux in a coil of magnetic flux density 12 units and area 7 units.
a) 0.84
b) 0.96
c) 8.4
d) 9.6

Answer: a [Reason:] The total flux in a coil is defined by φ = BA, where B = 12 and A = 0.07. On substituting these values, we get φ = 12 x 0.07 = 0.84 units.

4. Find the energy of a coil of inductance 18mH and current passing through it 1.25A.(in 10-3 order)
a) 14.06
b) 61
c) 46.1
d) 28.12

Answer: a [Reason:] The magnetic energy possessed by a coil is given by E = 0.5 x LI2. Put L = 18 x 10-3 and I = 1.25, thus we get E = 0.5 x 18 x 10-3 x 1.252 = 14.06 x 10-3 units.

5. Using Maxwell equation which of the following cannot be calculated directly?
a) B
b) D
c) A
d) H

Answer: c [Reason:] The Maxwell equations can be used to compute E,H,D,B and J directly. It is not possible to find the magnetic vector potential A directly.

6. Which of the following relation will hold good?
a) D = μ H
b) B = ε E
c) E = ε D
d) B = μ H

Answer: d [Reason:] The magnetic flux density is the product the permeability and the magnetic field intensity. This statement is always true for any material (permeability).

7. The permeability and permittivity of air or free space is unity. State true/false.
a) True
b) False

Answer: b [Reason:] The permeability and permittivity of free space or air is always unity. This implies that the air is always ready to store electric or magnetic charges subjected to it.

8. Choose the best relation.
a) A = -Div(V)
b) V = Curl(A)
c) H = -Grad(V)
d) V = Div(E)

Answer: c [Reason:] For any magnetic field, the magnetic field intensity will be the negative gradient of the potential of the field. This is given by H = -Grad(V).

9. Find the magnetic field when the magnetic vector potential is a unit vector.
a) 1
b) -1
c) 0
d) 2

Answer: c [Reason:] We know that H = -Grad(V), where is a unit vector. The gradient of a constant/unit vector will be zero. Thus the magnetic field intensity will be zero.

10. Find the force experienced by an electromagnetic wave in a conductor?
a) Electrostatic force
b) Magneto static force
c) Electro motive force
d) Lorentz force

Answer: d [Reason:] The electromagnetic wave experiences Lorentz force which is the combination of the electrostatic force and magneto static force. It is given by F = QE + Q(V X B).

## Set 5

1. Find the induced EMF in an inductor of 2mH and the current rate is 2000 units.
a) 4
b) -4
c) 1
d) -1

Answer: b [Reason:] The induced emf is given by e = -Ldi/dt. Put L = 2 x 10-3 and di/dt = 2000 in the equation. We get e = -2 x 10-3 x 2000 = -4 units.

2. Find the work done in an inductor of 4H when a current 8A is passed through it?
a) 256
b) 128
c) 64
d) 512

Answer: b [Reason:] The work done in the inductor will be W = 0.5 x LI2. On substituting L = 4 and I = 8, we get, W = 0.5 x 4 x 82 = 128 units.

3. Find the inductance of a material with 100 turns, area 12 units and current of 2A in air.
a) 0.75mH
b) 7.5mH
c) 75mH
d) 753mH

Answer: a [Reason:] The inductance of any material(coil) is given by L = μ N2A/I. On substituting N = 100, A = 0.12 and I = 2, we get L = 4π x 10-7 x 1002 x 0.12/2 = 0.75 units.

4. Calculate the magnetic energy when the magnetic intensity in air is given as 14.2 units(in 10-4 order)
a) 1.26
b) 2.61
c) 6.12
d) 1.62

Answer: a [Reason:] The magnetic energy is given by E = 0.5 μ H2. Put H = 14.2 and in air μ = 4π x 10-7, we get E = 0.5 x 4π x 10-7 x 14.22 = 1.26 x 10-4 units.

5. Calculate the magnetic energy when the magnetic flux density is given by 32 units(in 108 order)
a) 4.07
b) 7.4
c) 0.47
d) 7.04

Answer: a [Reason:] The magnetic energy is given by E = 0.5 μ H2 and we know that μH = B. On substituting we get a formula E = 0.5 B2/μ. Put B = 32 and in air μ = 4π x 10-7, we get E = 0.5 x 322/4π x 10-7 = 4.07 x 108 units.

6. Calculate the energy when the magnetic intensity and magnetic flux density are 15 and 65 respectively.
a) 755
b) 487.5
c) 922
d) 645

Answer: b [Reason:] The magnetic energy can also be written as E = 0.5 μH2 = 0.5 BH, since B = μH. On substituting B = 65 and H = 15 we get E = 0.5 x 65 x 15 = 487.5 units.

7. Find the inductance when the energy is given by 2 units with a current of 16A.
a) 15.6mH
b) 16.5mH
c) 16.8mH
d) 15.8mH

Answer: a [Reason:] The energy stored in an inductor is given by E = 0.5 LI2. To get L, put E = 2 and I = 16 and thus L = 2E/I2 = 2 x 2/162 = 15.6mH.

8. Find the power of an inductor of 5H and current 4.5A after 2 seconds.
a) 25.31
b) 50.62
c) 102.4
d) 204.8

Answer: a [Reason:] The energy stored in an inductor is given by E = 0.5 LI2. Thus, put L = 5 and I = 4.5 and we get E = 0.5 x 5 x 4.52 = 50.625 units To get power P = E/t = 50.625/2 = 25.31 units.

9. Find the turns in an solenoid of inductance 23.4mH , current 2A and area 15cm.
a) 900
b) 400
c) 498
d) 658