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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Electromagnetic Fields Theory (EMFT)  >  Test: Biot Savart Law - Electrical Engineering (EE) MCQ

Test: Biot Savart Law - Electrical Engineering (EE) MCQ


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

Test: Biot Savart Law for Electrical Engineering (EE) 2024 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) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Biot Savart Law below.
Solutions of Test: Biot Savart Law questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Biot Savart Law solutions in Hindi for Electromagnetic Fields Theory (EMFT) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Biot Savart Law | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Biot Savart Law - Question 1
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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
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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.

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Test: Biot Savart Law - Question 3
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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
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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
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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
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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
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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
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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
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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
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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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Biot Savart Law MCQs with Answers

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