This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Ampere Law”.

1. The point form of Ampere law is given by

a) Curl(B) = I

b) Curl(D) = J

c) Curl(V) = I

d) Curl(H) = J

View Answer

Explanation: Ampere law states that the line integral of H about any closed path is exactly equal to the direct current enclosed by that path. ∫ H.dl = I The point form will be Curl (H) = J.

2. The Ampere law is based on which theorem?

a) Green’s theorem

b) Gauss divergence theorem

c) Stoke’s theorem

d) Maxwell theorem

View Answer

Explanation: The proof of the Ampere’s circuital law is obtained from Stoke’s theorem for H and J only.

3. Electric field will be maximum outside the conductor and magnetic field will be maximum inside the conductor. State True/False.

a) True

b) False

View Answer

Explanation: At the conductor-free space boundary, electric field will be maximum and magnetic field will be minimum. This implies electric field is zero inside the conductor and increases as the radius increases and the magnetic field is zero outside the conductor and decreases as it approaches the conductor.

4. Find the magnetic flux density of a finite length conductor of radius 12cm and current 3A in air( in 10^{-6} order)

a) 4

b) 5

c) 6

d) 7

View Answer

Explanation: The magnetic field intensity is given by H = I/2πr, where I = 3A and r = 0.12. The magnetic flux density in air B = μ H, where μ = 4π x 10

^{-7}.Thus B = 4π x 10

^{-7}x 3/2π x 0.12 = 5x 10

^{-6}units.

5. Calculate the magnetic field intensity due to a toroid of turns 50, current 2A and radius 159mm.

a) 50

b) 75

c) 100

d) 200

View Answer

Explanation: The magnetic field intensity is given by H = NI/2πrm, where N = 50, I = 2A and rm = 1/2π. Thus H = 50 x 2/2π x 0.159 = 100 units.

6. Find the magnetic field intensity due to an infinite sheet of current 5A and charge density of 12j units in the positive y direction and the z component is above the sheet.

a) -6

b) 12k

c) 60

d) 6

View Answer

Explanation: The magnetic field intensity when the normal component is above the sheet is Hx = 0.5 K, where K = 12. Thus we get H = 0.5 x 12 = 6 units.

7. Find the magnetic field intensity due to an infinite sheet of current 5A and charge density of 12j units in the positive y direction and the z component is below the sheet.

a) 6

b) 0

c) -6

d) 60k

View Answer

Explanation: The magnetic intensity when the normal component is below the sheet is Hy = -0.5 K, where K = 12.Thus we get H = -0.5 x 12 = -6 units.

8. Find the current density on the conductor surface when a magnetic field H = 3cos x i + zcos x j A/m, for z>0 and zero, otherwise is applied to a perfectly conducting surface in xy plane.

a) cos x i

b) –cos x i

c) cos x j

d) –cos x j

View Answer

Explanation: By Ampere law, Curl (H) = J. The curl of H will be i(-cos x) – j(0) + k(-z sin x) = -cos x i – zsin x k. In the xy plane, z = 0. Thus Curl(H) = J = -cos x i.

9. When the rotational path of the magnetic field intensity is zero, then the current in the path will be

a) 1

b) 0

c) ∞

d) 0.5

View Answer

Explanation: By Ampere law, Curl(H) = J. The rotational path of H is zero, implies the curl of H is zero. This shows the current density J is also zero. The current is the product of the current density and area, which is also zero.

10. Find the magnetic field intensity when the current density is 0.5 units for an area up to 20 units.

a) 10

b) 5

c) 20

d) 40

View Answer

Explanation: We know that ∫ H.dl = I. By Stoke’s law, we can write Curl(H) = J. In integral form, H = ∫ J.ds, where J = 0.5 and ds is defined by 20 units. Thus H = 0.5 x 20 = 10 units.

**Sanfoundry Global Education & Learning Series – Electromagnetic Theory.**

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