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QUESTION: 1

The H quantity is analogous to which component in the following?

Solution:

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.

QUESTION: 2

The magnetic flux density is directly proportional to the magnetic field intensity. State True/False.

Solution:

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.

QUESTION: 3

Ampere law states that,

Solution:

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.

QUESTION: 4

Given the magnetic field is 2.4 units. Find the flux density in air(in 10^{-6} order).

Solution:

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.

QUESTION: 5

Find the electric field when the magnetic field is given by 2sin t in air.

Solution:

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.

QUESTION: 6

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.

Solution:

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.

QUESTION: 7

Find the magnetic field intensity due to a solenoid of length 12cm having 30 turns and current of 1.5A.

Solution:

Answer: d

Explanation: The magnetic field intensity of a solenoid is given by H = NI/L = 30 X 1.5/0.12 = 375 units.

QUESTION: 8

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.

Solution:

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.

QUESTION: 9

Find the magnetic field intensity of a toroid of turns 40 and radius 20cm. The current carried by the toroid be 3.25A.

Solution:

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.

QUESTION: 10

The magnetic field intensity of an infinite sheet of charge with charge density 36.5 units in air will be

Solution:

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.

QUESTION: 11

Identify which of the following is the unit of magnetic flux density?

Solution:

Answer: c

Explanation: The unit of magnetic flux density is weber/m^{2}. It is also called as tesla

QUESTION: 12

The divergence of H will be

Solution:

Answer: d

Explanation: We know that the divergence of B is zero. Also B = μH. Thus divergence of H is also zero.

QUESTION: 13

Find the flux contained by the material when the flux density is 11.7 Tesla and the area is 2 units.

Solution:

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.

QUESTION: 14

Find the current when the magnetic field intensity is given by 2L and L varies as 0->1.

Solution:

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.

QUESTION: 15

Find the magnetic field intensity when the flux density is 8 x 10^{-6} Tesla in the medium of air.

Solution:

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.

QUESTION: 16

If ∫ H.dL = 0, then which statement will be true?

Solution:

Answer: c

Explanation: The given condition shows that the magnetic field intensity will be the negative gradient of the magnetic vector potential.

QUESTION: 17

Find the magnetic flux density of the material with magnetic vector potential A = y i + z j + x k.

Solution:

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.

QUESTION: 18

Find the magnetic flux density when a flux of 28 units is enclosed in an area of 15cm.

Solution:

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.

QUESTION: 19

Find the magnetic flux density B when E is given by 3sin y i + 4cos z j + ex k.

Solution:

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.

QUESTION: 20

Find current density J when B = 50 x 10-6 units and area dS is 4 units.

Solution:

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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