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# Test: Ampere Law & Maxwell Law

## 20 Questions MCQ Test Electromagnetic Fields Theory | Test: Ampere Law & Maxwell Law

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

### The point form of Ampere law is given by

Solution:

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.

QUESTION: 2

### The Ampere law is based on which theorem?

Solution:

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

QUESTION: 3

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

Solution:

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.

QUESTION: 4

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

Solution:

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.

QUESTION: 5

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

Solution:

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.

QUESTION: 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.

Solution:

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.

QUESTION: 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.

Solution:

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.

QUESTION: 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.

Solution:

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.

QUESTION: 9

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

Solution:

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.

QUESTION: 10

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

Solution:

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.

QUESTION: 11

The divergence of which quantity will be zero?

Solution:

Explanation: The divergence of the magnetic flux density is always zero. This is because of the non existence of magnetic monopoles in a magnetic field.

QUESTION: 12

Find the charge density when the electric flux density is given by 2x i + 3y j + 4z k.

Solution:

Explanation: The charge density is the divergence of the electric flux density by Maxwell’s equation. Thus ρ = Div (D) and Div (D) = 2 + 3 + 4 = 9. We get ρ = 9 units.

QUESTION: 13

Find the Maxwell equation derived from Faraday’s law.

Solution:

Explanation: From the Faraday’s law and Lenz law, using Stoke’s theorem, we get Curl(E) = -dB/dt. This is the Maxwell’s first law of electromagnetics.

QUESTION: 14

Find the Maxwell law derived from Ampere law.

Solution:

Explanation: From the current density definition and Ohm’s law, the Ampere circuital law Curl(H) = J can be derived. This is Maxwell’s second law of electromagnetics.

QUESTION: 15

Solution:

Explanation: The stationary loop in a varying magnetic field results in an induced emf due to the change in the flux linkage of the loop. This emf is called as induced or transformer EMF.

QUESTION: 16

In which of the following forms can Maxwell’s equation not be represented?

Solution:

Explanation: Maxwell equations can be represented in differential/point form and integral form alternatively. Sometimes, it can be represented by time varying fields called harmonic form.

QUESTION: 17

The charge build up in the capacitor is due to which quantity?

Solution:

Explanation: The charge in the capacitor is due to displacement current. It is the current in the presence of the dielectric placed between two parallel metal plates.

QUESTION: 18

In metals which of the following equation will hold good?

Solution:

Explanation: Generally, the Curl(H) is the sum of two currents- conduction and displacement. In case of metals, it constitutes conduction J and in case of dielectrics, it constitutes the displacement current dD/dt.

QUESTION: 19

Find the flux enclosed by a material of flux density 12 units in an area of 80cm.

Solution:

Explanation: The total flux in a material is the product of the flux density and the area. It is given by flux = 12 x 0.8= 9.6 units.

QUESTION: 20

Find the electric flux density of a material with charge density 16 units in unit volume.

Solution: