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1.

Calculate the reluctance of a material with length 2π x 10^-4 in air with area 0.5.(a) 1(b) 10(c) 100(d) 1000The question was asked in an interview for internship.This key question is from Magnetic Energy and Circuits in division Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right choice is (d) 1000

The explanation: The reluctance is given by S = L/μ A, where L is the length, A is the area and μ is the PERMEABILITY. On substituting L = 2π x 10^-4, A = 0.5 and μ = 4π x 10^-7, we GET S = 10^3/(2×0.5) = 1000 units.

2.

Ampere turn is equivalent to which element?(a) Sφ(b) S/φ(c) φ/S(d) SThe question was posed to me by my college director while I was bunking the class.I'm obligated to ask this question of Magnetic Energy and Circuits in section Magnetic Forces and Materials of Electromagnetic Theory

Answer»
3.

Calculate the reluctance of the material with a mmf of 3.5 units and flux of7units.(a) 32.5(b) 10.5(c) 0.5(d) 2I had been asked this question during an internship interview.I'm obligated to ask this question of Magnetic Energy and Circuits in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer»

The correct OPTION is (c) 0.5

Best explanation: The reluctance is defined as the ratio of the mmf and the flux. It is GIVEN by S = mmf/φ. On SUBSTITUTING mmf = 3.5 and φ = 7, we get S = 3.5/7 = 0.5 units.

4.

The resistance in a magnetic material is called as(a) Capacitance(b) Inductance(c) Reluctance(d) Magnetic resistanceThis question was addressed to me in examination.The query is from Magnetic Energy and Circuits in division Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right OPTION is (c) RELUCTANCE

To elaborate: The reluctance of a magnetic material is the ABILITY of the material to oppose the magnetic flux. It is the ratio of the magnetic motive force mmf to the flux.

5.

The flux lines of two energised coils overlapping on each other will give(a) Series aiding(b) Shunt aiding(c) Series opposing(d) Shunt opposingThe question was asked in an interview for job.The above asked question is from Magnetic Energy and Circuits topic in division Magnetic Forces and Materials of Electromagnetic Theory

Answer»
6.

The energy in a magnetic material is due to which process?(a) Emf(b) Magnetization(c) Magnetostriction(d) PolarizationI had been asked this question by my college professor while I was bunking the class.This intriguing question comes from Magnetic Energy and Circuits topic in section Magnetic Forces and Materials of Electromagnetic Theory

Answer» CORRECT ANSWER is (b) Magnetization

Easiest EXPLANATION: The energy in a magnetic material is DUE to the formation of magnetic dipoles which are HELD together due to magnetic force. This gives energy to the material. Hence it is due to magnetization process.
7.

The induced emf in a material opposes the flux producing it. This is(a) Faraday law(b) Ampere law(c) Lenz law(d) Curie lawI have been asked this question in a job interview.Origin of the question is Magnetic Energy and Circuits in section Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right answer is (c) LENZ law

To explain I would say: The induced EMF in a material under the INFLUENCE of a MAGNETIC field will oppose the flux that PRODUCES it. This is indicated by a negative sign in the emf equation. This phenomenon is called Lenz law.

8.

The magnetic energy of a magnetic material is given by(a) BH/2(b) B/2H(c) H/2B(d) B/HThe question was asked during an internship interview.My enquiry is from Magnetic Energy and Circuits topic in section Magnetic Forces and Materials of Electromagnetic Theory

Answer»
9.

Find the inductance of a coil with turns 50, flux 3 units and a current of 0.5A(a) 150(b) 300(c) 450(d) 75I have been asked this question in semester exam.I'm obligated to ask this question of Inductances in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right choice is (B) 300

The explanation is: The self inductance of a coil is given by L = Nφ/I, where N = 50, φ = 3 and I = 0.5. Thus L = 50 X 3/0.5 = 300 UNITS.

10.

Calculate the mutual inductance of two tightly coupled coils with inductances 49H and 9H.(a) 21(b) 58(c) 40(d) 49/9I had been asked this question in an interview.My doubt is from Inductances topic in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Correct option is (a) 21

Easy explanation: For tightly coupled coils, the coefficient of coupling is UNITY. Then the MUTUAL INDUCTANCE will be M = √(L1 x L2)= √(49 x 9) = 21 units.

11.

The inductance is proportional to the ratio of flux to current. State True/False.(a) True(b) FalseThis question was addressed to me during an online exam.I would like to ask this question from Inductances topic in section Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Correct option is (a) True

To explain: The EXPRESSION is given by L = Ndφ/di. It can be seen that L is proportional to the RATIO of flux to CURRENT. Thus the statement is true.

12.

With unity coupling, the mutual inductance will be(a) L1 x L2(b) L1/L2(c) √(L1 x L2)(d) L2/L1The question was asked in an interview.My question is based upon Inductances topic in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer» CORRECT answer is (c) √(L1 x L2)

Easy EXPLANATION: The expression for MUTUAL inductance is given by M = K √(L1 x L2),where k is the coefficient of coupling. For unity coupling, k = 1, then M = √(L1 x L2).
13.

A coil is said to be loosely coupled with which of the following conditions?(a) K>1(b) K0.5(d) K

Answer»

Right choice is (d) K<0.5

The EXPLANATION is: k is the coefficient of coupling. It lies between 0 and 1. For loosely COUPLED COIL, the coefficient of coupling will be very less. THUS the condition K<0.5 is true.

14.

The equivalent inductance of two coils with series opposing flux having inductances 7H and 2H with a mutual inductance of 1H.(a) 10(b) 7(c) 11(d) 13I had been asked this question in a job interview.I would like to ask this question from Inductances in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer»

The correct option is (B) 7

The BEST I can explain: The equivalent inductance of TWO coils in series with opposing FLUX is L = L1 + L2 – 2M, where L1 and L2 are the self inductances and M is the mutual inductance. Thus L = 7 + 2 – 2(1) = 7H.

15.

The expression for the inductance in terms of turns, flux and current is given by(a) L = N dφ/di(b) L = -N dφ/di(c) L = Niφ(d) L = Nφ/iI got this question in exam.The doubt is from Inductances topic in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right CHOICE is (a) L = N dφ/di

The best EXPLANATION: We know that e = -N dφ/dt and also e = -L di/dt. On EQUATING both we GET, L = Ndφ/di is the expression for inductance.

16.

The equivalent inductances of two coils 2H and 5H in series aiding flux with mutual inductance of 3H is(a) 10(b) 30(c) 1(d) 13I got this question during an internship interview.My question is taken from Inductances topic in section Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right option is (d) 13

To EXPLAIN: The EQUIVALENT INDUCTANCE of two coils in series is given by L = L1 + L2 + 2M, where L1 and L2 are the self inductances and M is the mutual inductance. THUS L = 2 + 5 + 2(3) = 13H.

17.

Find the ratio of permeability of the two media when the wave is incident on the boundary at 45 degree and reflected by the boundary at 60 degree.(a) 1:1(b) √3:1(c) 1:√3(d) 1:√2This question was posed to me by my college professor while I was bunking the class.This intriguing question originated from Magnetic Boundary Conditions topic in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer»

The correct option is (c) 1:√3

Explanation: From the MAGNETIC boundary conditions, the ratio of PERMEABILITY μ1/μ2 = tan θ1/tan θ2 and θ1 = 45, θ2 = 60. THUS we GET μ1/μ2 = 1/√3. The ratio will be 1:√3.

18.

Calculate the emf of a coil with turns 100 and flux rate 5 units.(a) 20(b) -20(c) 500(d) -500I had been asked this question in an interview.I want to ask this question from Inductances in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Correct option is (d) -500

To elaborate: The emf is the product of the turns of the coil and the FLUX RATE. Thus e = -N dφ/dt, where the negative SIGN indicates that the emf induced is OPPOSING the flux. Thus e = -100 X 5 = -500 units.

19.

Find the magnetic moment of a material with magnetization 5 units in a volume of 35 units.(a) 7(b) 1/7(c) 15(d) 175I got this question during an interview for a job.This key question is from Magnetic Boundary Conditions in division Magnetic Forces and Materials of Electromagnetic Theory

Answer» CORRECT choice is (d) 175

For EXPLANATION I would say: The magnetization is the ratio of the MAGNETIC moment and the volume. To GET moment, put M = 5 and V = 35, THUS moment will be 5 x 35 = 175 units.
20.

Find the magnetization of the material with susceptibility of 6 units and magnetic field intensity of 13 units.(a) 2.16(b) 6.2(c) 78(d) 1.3The question was posed to me during a job interview.Asked question is from Magnetic Boundary Conditions topic in division Magnetic Forces and Materials of Electromagnetic Theory

Answer» CORRECT answer is (c) 78

Explanation: The magnetization is the PRODUCT of the susceptibility and the magnetic field intensity. THUS M = 6 X 13 = 78 units.
21.

The line integral of the magnetic field intensity is the(a) Current density(b) Current(c) Magnetic flux density(d) Magnetic momentThis question was addressed to me in my homework.Query is from Magnetic Boundary Conditions in division Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right option is (b) CURRENT

The explanation is: The LINE INTEGRAL of the magnetic FIELD intensity is given by ∫H.dl. This is same as the current COMPONENT. From this relation, the Ampere law can be deduced.

22.

The normal component of magnetic field intensity at the boundary of separation of the medium will be(a) Same(b) Different(c) Negative(d) InverseI have been asked this question during an online exam.I need to ask this question from Magnetic Boundary Conditions in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Correct OPTION is (a) Same

Easy explanation: The normal COMPONENT and tangential components of the MAGNETIC flux DENSITY will be same. This holds good for any medium.

23.

The flux density of medium 1 has a normal component of 2.4 units, then the normal component of the flux density in the medium 2 will be(a) 1.2(b) 4.8(c) 2.4(d) 0I have been asked this question in final exam.This interesting question is from Magnetic Boundary Conditions topic in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer»

The correct CHOICE is (c) 2.4

The BEST I can explain: Unlike the ELECTRIC fields, the magnetic flux DENSITY has normal component same in both the MEDIUMS. This gives Bn1 = Bn2.

24.

In air, the tangential component of flux density is continuous at the boundary. State True/False.(a) True(b) FalseThe question was posed to me in an interview.This interesting question is from Magnetic Boundary Conditions in division Magnetic Forces and Materials of Electromagnetic Theory

Answer»

The correct option is (a) True

The best explanation: SINCE the TANGENTIAL component of the magnetic field intensity will be continuous and B = μH, in air, the tangential component of the FLUX density will ALSO be continuous.

25.

The tangential component of the magnetic field intensity is continuous at the boundary of separation of two media. State True/False.(a) True(b) FalseI had been asked this question in an interview for job.My doubt is from Magnetic Boundary Conditions topic in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Correct choice is (a) True

For explanation I WOULD say: For two medium of separation, the tangential component of the MAGNETIC field INTENSITY will be CONTINUOUS. This is analogous to the fact that the tangential component of the electric field intensity is continuous at the boundary.

26.

Find the correct relation between current density and magnetization.(a) J = Grad(M)(b) J = Div(M)(c) J = Curl(M)(d) M = Curl(J)The question was asked in an interview.This intriguing question comes from Magnetic Boundary Conditions in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer»

The correct choice is (C) J = Curl(M)

Easiest explanation: The curl of the MAGNETIZATION gives the magnetic field intensity theoretically. From MAXWELL equation, we can correlate that with the CURRENT density (Ampere law)

27.

Which of the following materials is ferrimagnetic?(a) Fe(b) Sn(c) Fe2O3(d) FeClThe question was posed to me in a job interview.Question is from Magnetization in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer» CORRECT answer is (c) Fe2O3

The best I can explain: 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.
28.

Very small and positive susceptibility is found in(a) Ferromagnetic(b) Diamagnetic(c) Paramagnetic(d) AntiferromagneticThis question was addressed to me in class test.My query is from Magnetization topic in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Correct choice is (c) Paramagnetic

Easiest explanation: Paramagnetic materials are CHARACTERIZED by a small and POSITIVE SUSCEPTIBILITY. The susceptibility and the temperature are DIRECTLY RELATED.

29.

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) 37.5I had been asked this question in examination.This interesting question is from Magnetization topic in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right choice is (a) 12.5

Easy EXPLANATION: 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.

30.

Find the permeability of a medium whose susceptibility is 100.(a) -100(b) 99(c) -99(d) 101I have been asked this question during an interview for a job.My question is taken from Magnetization topic in division Magnetic Forces and Materials of Electromagnetic Theory

Answer» RIGHT answer is (d) 101

For explanation: The susceptibility is given by χm = μr-1. To get permeability, μr = χm + 1 = 100 + 1 = 101 UNITS.
31.

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/IAVThis question was addressed to me in exam.The doubt is from Magnetization in division Magnetic Forces and Materials of Electromagnetic Theory

Answer»
32.

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) 2.34I have been asked this question during an interview.My enquiry is from Magnetization in section Magnetic Forces and Materials of Electromagnetic Theory

Answer» CORRECT answer is (a) 1.78

Easiest EXPLANATION: 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.
33.

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 θThe question was posed to me in unit test.Query is from Magnetization in section Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right ANSWER is (c) T = BIA sin θ

Easy explanation: 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.

34.

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) 1I had been asked this question during an interview for a job.Origin of the question is Magnetization in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer» RIGHT choice is (a) 10

The explanation is: 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.
35.

The susceptibility is independent of temperature in which material?(a) Paramagnetic(b) Ferromagnetic(c) Diamagnetic(d) FerromagneticThis question was posed to me during an interview.I'm obligated to ask this question of Magnetic Materials topic in section Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Correct choice is (c) Diamagnetic

Explanation: 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.

36.

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) 45I got this question during an online interview.Question is from Magnetization topic in division Magnetic Forces and Materials of Electromagnetic Theory

Answer»

The correct answer is (b) 36

Easy explanation: 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.

37.

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) 200This question was addressed to me in an interview for internship.I'm obligated to ask this question of Magnetic Materials topic in chapter Magnetic Forces and Materials of Electromagnetic Theory

Answer»

The correct CHOICE is (b) 20

The best EXPLANATION: 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.

38.

The materials having very small susceptibility at all temperatures are(a) Antiferromagnetic(b) Diamagnetic(c) Ferromagnetic(d) ParamagneticThe question was posed to me during an interview for a job.This interesting question is from Magnetic Materials in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Correct option is (a) Antiferromagnetic

The best I can explain: In antiferromagnetic materials, the SUSCEPTIBILITY will decrease with increase in TEMPERATURE. They have RELATIVELY SMALL susceptibility at all temperatures.

39.

The converse of magnetostriction is called the(a) Magnetization(b) Magnetic anisotropy(c) Villari effect(d) Curie effectThis question was addressed to me by my school teacher while I was bunking the class.The doubt is from Magnetic Materials in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right choice is (C) Villari effect

For explanation I would SAY: 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.

40.

In which materials the magnetic anisotropy is followed?(a) Diamagnetic(b) Paramagnetic(c) Ferromagnetic(d) FerromagneticI had been asked this question during an online interview.This intriguing question originated from Magnetic Materials topic in section Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right answer is (c) Ferromagnetic

The best I can explain: 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.

41.

Piezoelectric effect is analogous to which phenomenon?(a) Electrostriction(b) Magnetostriction(c) Anisotropy(d) MagnetizationThe question was posed to me by my college professor while I was bunking the class.Query is from Magnetic Materials in division Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right choice is (b) Magnetostriction

Best explanation: 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.

42.

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) 19.4The question was posed to me during an interview.I want to ask this question from Magnetic Materials in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right CHOICE is (d) 19.4

To explain I would say: 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.

43.

The magnetic materials follow which law?(a) Faraday’s law(b) Ampere law(c) Lenz law(d) Curie Weiss lawThis question was posed to me in a job interview.My question is taken from Magnetic Materials topic in section Magnetic Forces and Materials of Electromagnetic Theory

Answer» CORRECT OPTION is (d) Curie WEISS law

To elaborate: Generally, the FERROMAGNETIC, paramagnetic and diamagnetic materials follow the Curie Weiss law, which relates the magnetization and the applied FIELD.
44.

The presence of parallel alignment of magnetic dipole moment is given by which materials?(a) Diamagnetic(b) Ferromagnetic(c) Paramagnetic(d) FerromagneticThis question was posed to me in final exam.My doubt stems from Magnetic Materials in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer» CORRECT option is (b) Ferromagnetic

To ELABORATE: The ferromagnetic materials are characterized by parallel ALIGNMENT of magnetic dipole moments. Their SUSCEPTIBILITY is very large.
45.

Find the magnetization of the field which has a magnetic moment 16 units in a volume of 1.2 units.(a) 16.67(b) 13.33(c) 15.56(d) 18.87I have been asked this question during a job interview.This intriguing question comes from Magnetic Dipole in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer»

The CORRECT OPTION is (b) 13.33

The explanation: The magnetization is the ratio of the MAGNETIC moment to the volume. Thus M = m/v, where m = 16 and v = 1.2. We get M = 16/1.2 = 13.33 units.

46.

The Bohr magneton is given by(a) eh/2m(b) eh/2πm(c) eh/4m(d) eh/4πmThe question was asked by my school principal while I was bunking the class.Question is taken from Magnetic Dipole topic in division Magnetic Forces and Materials of Electromagnetic Theory

Answer»

The correct choice is (d) eh/4πm

The best I can explain: In atomic physics, the Bohr magneton (symbol μB) is a PHYSICAL CONSTANT and the natural unit for expressing the magnetic moment of an ELECTRON caused by either its orbital or spin angular momentum. It is GIVEN by eh/4πm, where H is the Planck’s constant, e is the charge of the electron and m is the mass of the electron.

47.

Calculate the Larmer angular frequency for a magnetic flux density of 12.34 x 10^-10.(a) 108.36(b) 810.63(c) 368.81(d) 183.36The question was posed to me in a job interview.I'd like to ask this question from Magnetic Dipole in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer»
48.

The ratio of the orbital dipole moment to the orbital angular moment is given by(a) e/m(b) –e/m(c) e/2m(d) –e/2mI got this question in an online quiz.This question is from Magnetic Dipole in division Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Right choice is (d) –e/2m

To explain I would say: The ORBITAL dipole moment is GIVEN by M = 0.5 x eVangx r^2 and the orbital angular moment is given by Ma = m x Vangx r^2. Their ratio M/Ma is given by –e/2m, the negative SIGN indicates the CHARGE of electron.

49.

Find the orbital angular moment of a dipole with angular velocity of 1.6m/s and radius 35cm(in 10-31 order)(a) 1.78(b) 8.71(c) 7.18(d) 2.43I got this question during an internship interview.The query is from Magnetic Dipole topic in portion Magnetic Forces and Materials of Electromagnetic Theory

Answer» CORRECT option is (a) 1.78

Best EXPLANATION: The orbital angular moment is given by Ma = m X Vangx R^2,where m = 9.1 x 10^-31, Vang = 1.6 and r = 0.35. On SUBSTITUTING, we get, Ma = 9.1 x 10^-31 x 1.6 x 0.35^2 = 1.78 x 10^-31 units.
50.

Find the orbital dipole moment in a field of dipoles of radius 20cm and angular velocity of 2m/s(in 10^-22 order)(a) 64(b) 76(c) 54(d) 78The question was asked by my college professor while I was bunking the class.This intriguing question originated from Magnetic Dipole in section Magnetic Forces and Materials of Electromagnetic Theory

Answer»

Correct ANSWER is (a) 64

To explain: The orbital dipole moment is given by M = 0.5 X eVangx R^2, where e = 1.6 x 10^-19 is the CHARGE of the electron, Vang = 2 and r = 0.2. On substituting, we get M = 0.5 x 1.6 x 10^-19x 2 x 0.2^2= 64 x 10^-22 units.