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Find the force that exists in an electromagnetic wave.
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.
In an field having a force of 12N and distance 20cm, the torque will be
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.
Find the torque in a conductor having current 2A, flux density 50 units, length 15cm and distance of 8m.
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.
The distance of the conductor when the area and length of the conductor is 24m2 and 13.56m.
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.
The torque on a conductor with flux density 23 units, current 1.6A and area 6.75 units will be
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.
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
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.
The torque of a conductor is defined only in the case when
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.
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.
The magnetic moment and torque are related as follows
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.
Calculate the magnetic moment when a field of B= 51 units is subjected to a torque of 20 units.
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.
The magnetic moment of a field with current 12A and area 1.6 units is
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.
Find the torque of a loop with magnetic moment 12.5 and magnetic flux density 7.65 units is
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.
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.
Find the orbital dipole moment in a field of dipoles of radius 20cm and angular velocity of 2m/s(in 10^{22} order)
Answer: a
Explanation: 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^{19}x 2 x 0.2^{2}= 64 x 10^{22} units.
Find the orbital angular moment of a dipole with angular velocity of 1.6m/s and radius 35cm(in 1031 order)
Answer: a
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.
The ratio of the orbital dipole moment to the orbital angular moment is given by
Answer: d
Explanation: 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.
Calculate the Larmer angular frequency for a magnetic flux density of 12.34 x 10^{10}.
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.
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.
Find the magnetization of the field which has a magnetic moment 16 units in a volume of 1.2 units.
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.
Which of the following is true regarding magnetic lines of force?
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.
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