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The H quantity is analogous to which component in the following?
Answer: c
Explanation: The H quantity refers to magnetic field intensity in the magnetic field. This is analogous to the electric field intensity E in the electric field.
The magnetic flux density is directly proportional to the magnetic field intensity. State True/False.
Answer: a
Explanation: The magnetic field intensity is directly proportional to the magnetic field intensity for a particular material (Permeability). It is given by B = μH.
Answer: d
Explanation: Ampere circuital law or Ampere law states that the closed integral of the magnetic field intensity is same as the current enclosed by it. It is given by Curl(H) = J.
Given the magnetic field is 2.4 units. Find the flux density in air(in 10^{6} order).
Answer: b
Explanation: We know that B = μH. On substituting μ = 4π x 10^{7} and H = 2.4, we get B = 4π x 10^{7} x 2.4 = 3 x 10^{6} units.
Find the electric field when the magnetic field is given by 2sin t in air.
Answer: a
Explanation: Given H = 2sin t. We get B = μH = 4π x 10^{7} x 2sin t = 8πx10^{7}sin t.
To get E, integrate B with respect to time, we get 8πx10^{7}cos t.
Find the height of an infinitely long conductor from point P which is carrying current of 6.28A and field intensity is 0.5 units.
Answer: b
Explanation: The magnetic field intensity of an infinitely long conductor is given by H = I/2πh. Put I = 6.28 and H = 0.5, we get h = 1/0.5 = 2 units.
Find the magnetic field intensity due to a solenoid of length 12cm having 30 turns and current of 1.5A.
Answer: d
Explanation: The magnetic field intensity of a solenoid is given by H = NI/L = 30 X 1.5/0.12 = 375 units.
Find the magnetic field intensity at the radius of 6cm of a coaxial cable with inner and outer radii are 1.5cm and 4cm respectively. The current flowing is 2A.
Answer: c
Explanation: The inner radius is 1.5cm and the outer radius is 4cm. It is clear that the magnetic field intensity needs to be calculated outside of the conductor ie, r>4cm. This will lead to zero, since H outside the conductor will be zero.
Find the magnetic field intensity of a toroid of turns 40 and radius 20cm. The current carried by the toroid be 3.25A.
Answer: a
Explanation: The magnetic field intensity of a toroid is given by H = NI/2πrm. Put N = 40, I = 3.25 and rm = 0.2, we get H = 40 x 3.25/2π x 0.2 = 103.45 units.
The magnetic field intensity of an infinite sheet of charge with charge density 36.5 units in air will be
Answer: a
Explanation: The magnetic field intensity of an infinite sheet of charge is given by H = 0.5 K, for the point above the sheet and –0.5 K, for the point below the sheet. Here k is the charge density. Thus H = 0.5 x 36.5 = 18.25 units.
Identify which of the following is the unit of magnetic flux density?
Answer: c
Explanation: The unit of magnetic flux density is weber/m^{2}. It is also called as tesla
Answer: d
Explanation: We know that the divergence of B is zero. Also B = μH. Thus divergence of H is also zero.
Find the flux contained by the material when the flux density is 11.7 Tesla and the area is 2 units.
Answer: a
Explanation: The total flux is given by φ = ∫ B.ds, where ∫ds is the area. Thus φ = BA. We get φ = 11.7 x 2 = 23.4 units.
Find the current when the magnetic field intensity is given by 2L and L varies as 0>1.
Answer: b
Explanation: From Ampere law, we get ∫ H.dL = I. Put H = 2L and L = 0>1. On integrating H with respect to L, the current will be 1A.
Find the magnetic field intensity when the flux density is 8 x 10^{6} Tesla in the medium of air.
Answer: a
Explanation: We how that, B = μH. To get H = B/μ, put B = 8 x 10^{6} and μ = 4π x 10^{7}. Thus H = 8 x 10^{6}/ 4π x 10^{7} = 6.36 units.
If ∫ H.dL = 0, then which statement will be true?
Answer: c
Explanation: The given condition shows that the magnetic field intensity will be the negative gradient of the magnetic vector potential.
Find the magnetic flux density of the material with magnetic vector potential A = y i + z j + x k.
Answer: b
Explanation: The magnetic flux density is the curl of the magnetic vector potential. B = Curl(A). Thus Curl(A) = i(1) – j(1) + k(1) = i – j – k. We get B = i – j – k.
Find the magnetic flux density when a flux of 28 units is enclosed in an area of 15cm.
Answer: b
Explanation: The total flux is the product of the magnetic flux density and the area. Total flux = B x A. To get B, put flux/area. B = 28/0.15 = 186.67 units.
Find the magnetic flux density B when E is given by 3sin y i + 4cos z j + ex k.
Answer: b
Explanation: We know that Curl (E) = dB/dt. The curl of E is (4sin z i – ex j – 3cos y k). To get B, integrate the curl(E) with respect to time to get B = ∫(4sin z i – ex j – 3cos y k)dt.
Find current density J when B = 50 x 106 units and area dS is 4 units.
Answer: a
Explanation: To get H, H = B/μ = 50 x 10^{6}/ 4π x 10^{7} = 39.78 units. Also H = ∫ J.dS, where H = 39.78 and ∫ dS = 4. Thus J = 39.78/4 = 9.94 units.
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