Test: Ampere Law & Maxwell Law

# Test: Ampere Law & Maxwell Law

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

Test: Ampere Law & Maxwell Law for Electrical Engineering (EE) 2022 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Ampere Law & Maxwell Law questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Ampere Law & Maxwell Law MCQs are made for Electrical Engineering (EE) 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Ampere Law & Maxwell Law below.
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Test: Ampere Law & Maxwell Law - Question 1

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 1

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.

Test: Ampere Law & Maxwell Law - Question 2

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 2

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

Test: Ampere Law & Maxwell Law - Question 3

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 3

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.

Test: Ampere Law & Maxwell Law - Question 4

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 4

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.

Test: Ampere Law & Maxwell Law - Question 5

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 5

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.

Test: Ampere Law & Maxwell Law - 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.

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 6

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.

Test: Ampere Law & Maxwell Law - 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.

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 7

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.

Test: Ampere Law & Maxwell Law - 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.

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 8

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.

Test: Ampere Law & Maxwell Law - Question 9

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 9

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.

Test: Ampere Law & Maxwell Law - Question 10

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 10

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.

Test: Ampere Law & Maxwell Law - Question 11

The divergence of which quantity will be zero?

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 11

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.

Test: Ampere Law & Maxwell Law - Question 12

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 12

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.

Test: Ampere Law & Maxwell Law - Question 13

Find the Maxwell equation derived from Faraday’s law.

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 13

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.

Test: Ampere Law & Maxwell Law - Question 14

Find the Maxwell law derived from Ampere law.

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 14

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.

Test: Ampere Law & Maxwell Law - Question 15

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 15

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.

Test: Ampere Law & Maxwell Law - Question 16

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 16

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.

Test: Ampere Law & Maxwell Law - Question 17

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 17

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.

Test: Ampere Law & Maxwell Law - Question 18

In metals which of the following equation will hold good?

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 18

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.

Test: Ampere Law & Maxwell Law - Question 19

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 19

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.

Test: Ampere Law & Maxwell Law - Question 20

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

Detailed Solution for Test: Ampere Law & Maxwell Law - Question 20

Explanation: The electric flux density from Maxwell’s equation is given by D = ∫ ρ dv. On substituting ρ = 16 and ∫dv = 1, we get D = 16 units.

## Electromagnetic Fields Theory (EMFT)

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## Electromagnetic Fields Theory (EMFT)

11 videos|46 docs|62 tests