Test: Magnetic Torque And Dipole


20 Questions MCQ Test Electromagnetic Theory | Test: Magnetic Torque And Dipole


Description
This mock test of Test: Magnetic Torque And Dipole for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 20 Multiple Choice Questions for Electrical Engineering (EE) Test: Magnetic Torque And Dipole (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Magnetic Torque And Dipole quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Magnetic Torque And Dipole exercise for a better result in the exam. You can find other Test: Magnetic Torque And Dipole extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

Find the force that exists in an electromagnetic wave.

Solution:

Answer: c
Explanation: In an electromagnetic wave, the force of the electric and magnetic field both coexist. This is given by F = qE + q(v x B). It is called Lorentz force.

QUESTION: 2

In an field having a force of 12N and distance 20cm, the torque will be

Solution:

Answer: b
Explanation: The torque is defined as the product of the force and distance in a field. Thus T = F x d = 12 x 0.2 = 2.4 units.

QUESTION: 3

Find the torque in a conductor having current 2A, flux density 50 units, length 15cm and distance of 8m.

Solution:

Answer: a
Explanation: The torque on a conductor is given by T = BILd, where L x d is the area of the conductor. Thus the torque will be, T = 50 x 2 x 0.15 x 8 = 120 units.

QUESTION: 4

The distance of the conductor when the area and length of the conductor is 24m2 and 13.56m.

Solution:

Answer: a
Explanation: We know that the surface integral is the area component which is the product of two dimensions given by length and distance in a conductor. Thus A = L x d. To get d, d = A/L = 24/13.56 = 1.76 units.

QUESTION: 5

The torque on a conductor with flux density 23 units, current 1.6A and area 6.75 units will be

Solution:

Answer: a
Explanation: The maximum torque on a conductor will be at perpendicular angle ie, at 90. The torque will be given as T = BIA, where B = 23, I = 1.6 and A = 6.75.Thus we get, T = 23 x 1.6 x 6.75 = 248.4 units.

QUESTION: 6

Consider the conductor to be a coil of turns 60 and the flux density to be 13.5 units, current 0.12A and area 16units. The torque will be

Solution:

Answer: a
Explanation: For a single turn or loop, the torque will be BIA. For N turns, the torque will be T = NBIA, where N = 60, B = 13.5, I = 0.12 and A = 16. Thus T = 60 x 13.5 x 0.12 x 16 = 1555.2 units.

QUESTION: 7

The torque of a conductor is defined only in the case when

Solution:

Answer: b
Explanation: The torque of a conductor is given by T = NBIA. This equation of the conductor is valid only when the plane of the loop is parallel to the magnetic field applied to it.

QUESTION: 8

Find the angle at which the torque is minimum.

Solution:

Answer: d
Explanation: The torque of a conductor loop is given by T = BIA cos θ. The torque is minimum refers to zero torque. This is possible only when the angle is 90 or perpendicular.

QUESTION: 9

The magnetic moment and torque are related as follows

Solution:

Answer: a
Explanation: The torque is defined as the product of the magnetic flux density and the magnetic moment. It is given by T = BM, where M = IA is the magnetic moment.

QUESTION: 10

Calculate the magnetic moment when a field of B= 51 units is subjected to a torque of 20 units.

Solution:

Answer: a
Explanation: The magnetic moment is given by the ratio of the torque and the magnetic flux density. Thus M = T/B, where T = 20 and B = 51 units. We get M = 20/51 = 0.39 units.

QUESTION: 11

The magnetic moment of a field with current 12A and area 1.6 units is

Solution:

Answer: a
Explanation: The magnetic moment is the product of current and the area of the conductor. It is given by M = IA, where I = 12 and A = 1.6.Thus we get, M = 12 x 1.6 = 19.2 units.

QUESTION: 12

Find the torque of a loop with magnetic moment 12.5 and magnetic flux density 7.65 units is

Solution:

Answer: a
Explanation: The torque is defined as the product of the magnetic moment and the magnetic flux density given by T = MB, where M = 12.5 and B = 7.65. Thus we get T = 12.5 x 7.65 = 95.625 units.

QUESTION: 13

The magnetization is defined by the ratio of

Solution:

Answer: b
Explanation: The magnetization refers to the amount of dipole formation in a given volume when it is subjected to a magnetic field. It is given by the ratio of the magnetic moment to the volume. Thus Pm = M/V.

QUESTION: 14

Find the orbital dipole moment in a field of dipoles of radius 20cm and angular velocity of 2m/s(in 10-22 order)

Solution:

Answer: a
Explanation: The orbital dipole moment is given by M = 0.5 x eVangx r2, 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 0.22= 64 x 10-22 units.

QUESTION: 15

Find the orbital angular moment of a dipole with angular velocity of 1.6m/s and radius 35cm(in 10-31 order)

Solution:

Answer: a
Explanation: The orbital angular moment is given by Ma = m x Vangx r2,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 0.352 = 1.78 x 10-31 units.

QUESTION: 16

The ratio of the orbital dipole moment to the orbital angular moment is given by

Solution:

Answer: d
Explanation: The orbital dipole moment is given by M = 0.5 x eVangx r2 and the orbital angular moment is given by Ma = m x Vangx r2. Their ratio M/Ma is given by –e/2m, the negative sign indicates the charge of electron.

QUESTION: 17

Calculate the Larmer angular frequency for a magnetic flux density of 12.34 x 10-10.

Solution:

Answer: a
Explanation: The Larmer angular frequency is the product of magnitude of the ratio of orbital dipole moment to orbital angular moment and the magnetic flux density. It is given by fL = B e/2m, where is the charge of electron and m is the mass of the electron. On substituting, we get fL = 12.34 x 10-10 x 1.6 x 10-19/(2 x 9.1 x 10-31) = 108.36 units.

QUESTION: 18

The Bohr magneton is given by

Solution:

Answer: d
Explanation: 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.

QUESTION: 19

Find the magnetization of the field which has a magnetic moment 16 units in a volume of 1.2 units.

Solution:

Answer: b
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.

QUESTION: 20

Which of the following is true regarding magnetic lines of force?

Solution:

Answer: b
Explanation: Magnetic Lines of Force is a an imaginary line representing the direction of magnetic field such that the tangent at any point is the direction of the field vector at that point.

Related tests