Test: Biot Savart Law


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10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Biot Savart Law

Test: Biot Savart Law for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Biot Savart Law questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Biot Savart Law MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Biot Savart Law below.
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Test: Biot Savart Law - Question 1

Biot Savart law in magnetic field is analogous to which law in electric field?

Detailed Solution for Test: Biot Savart Law - Question 1

Answer: c
Explanation: Biot Savart law states that the magnetic flux density H = I.dl sinθ/4πr2, which is analogous to the electric field F = q1q2/4πεr2, which is the Coulomb’s law.

Test: Biot Savart Law - Question 2

Which of the following cannot be computed using the Biot Savart law?

Detailed Solution for Test: Biot Savart Law - Question 2

Answer: c
Explanation: The Biot Savart law is used to calculate magnetic field intensity. Using which we can calculate flux density and permeability by the formula B = μH.

Test: Biot Savart Law - Question 3

Find the magnetic field of a finite current element with 2A current and height 1/2π is

Detailed Solution for Test: Biot Savart Law - Question 3

Answer: a
Explanation: The magnetic field due to a finite current element is given by H = I/2πh. Put I = 2 and h = 1/2π, we get H = 1 unit.

Test: Biot Savart Law - Question 4

Calculate the magnetic field at a point on the centre of the circular conductor of radius 2m with current 8A.

Detailed Solution for Test: Biot Savart Law - Question 4

Answer: b
Explanation: The magnetic field due to a point in the centre of the circular conductor is given by H = I/2a. Put I = 8A and a = 2m, we get H = 8/4 = 2 units.

Test: Biot Savart Law - Question 5

The current element of the solenoid of turns 100, length 2m and current 0.5A is given by,

Detailed Solution for Test: Biot Savart Law - Question 5

Answer: c
Explanation: The current element of the solenoid is given by NI dx/L. Put N = 100, I = 0.5 and L = 2 to get, I dx = 100 x 0.5 x dx/2 = 25 dx.

Test: Biot Savart Law - Question 6

Find the magnetic field intensity at the centre O of a square of the sides equal to 5m and carrying 10A of current.

Detailed Solution for Test: Biot Savart Law - Question 6

Answer: d
Explanation: The magnetic field is given by H = 4I/√2πω. Put I = 10 and ω = 5m. Thus H = 4 x 10/√2π(5) = 1.8 unit.

Test: Biot Savart Law - Question 7

Find the magnetic flux density when a point from a finite current length element of current 0.5A and radius 100nm.

Detailed Solution for Test: Biot Savart Law - Question 7

Answer: c
Explanation: The magnetic flux density is B = μH, where H is given by I/2πr. Put μ = 4π x 10-7, I = 0.5 and r = 10-7, we get B = 4π x 10-7 x 0.5/2π x 10-7 = 1 unit

Test: Biot Savart Law - Question 8

In a static magnetic field only magnetic dipoles exist. State True/False. 

Detailed Solution for Test: Biot Savart Law - Question 8

Answer: a
Explanation: From Gauss law for magnetic field, we get divergence of the magnetic flux density is always zero (ie, Div(B) = 0). This implies the non-existence of magnetic monopole.

Test: Biot Savart Law - Question 9

The magnetic field intensity will be zero inside a conductor. State true/false.

Detailed Solution for Test: Biot Savart Law - Question 9

Answer: b
Explanation: Electric field will be zero inside a conductor and magnetic field will be zero outside the conductor. In other words, the conductor boundary, E will be maximum and H will be minimum.

Test: Biot Savart Law - Question 10

Find the magnetic field when a circular conductor of very high radius is subjected to a current of 12A and the point P is at the centre of the conductor.

Detailed Solution for Test: Biot Savart Law - Question 10

Answer: c
Explanation: The magnetic field of a circular conductor with point on the centre is given by I/2a. If the radius is assumed to be infinite, then H = 12/2(∞) = 0.

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