Electromagnetic Theory Questions and Answers – Biot Savart Law

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

1. Biot Savart law in magnetic field is analogous to which law in electric field?
a) Gauss law
b) Faraday law
c) Coulomb’s law
d) Ampere law
View Answer

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.
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2. Which of the following cannot be computed using the Biot Savart law?
a) Magnetic field intensity
b) Magnetic flux density
c) Electric field intensity
d) Permeability
View Answer

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.

3. Find the magnetic field of a finite current element with 2A current and height 1/2π is
a) 1
b) 2
c) 1/2
d) 1/4
View Answer

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.
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4. Calculate the magnetic field at a point on the centre of the circular conductor of radius 2m with current 8A.
a) 1
b) 2
c) 3
d) 4
View Answer

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.

5. The current element of the solenoid of turns 100, length 2m and current 0.5A is given by,
a) 100 dx
b) 200 dx
c) 25 dx
d) 50 dx
View Answer

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.
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6. Find the magnetic field intensity at the centre O of a square of the sides equal to 5m and carrying 10A of current.
a) 1.2
b) 1
c) 1.6
d) 1.8
View Answer

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.

7. Find the magnetic flux density when a point from a finite current length element of current 0.5A and radius 100nm.
a) 0
b) 0.5
c) 1
d) 2
View Answer

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.
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8. In a static magnetic field only magnetic dipoles exist. State True/False.
a) True
b) False
View Answer

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.

9. The magnetic field intensity will be zero inside a conductor. State true/false.
a) True
b) False
View Answer

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.
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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.
a) 1
b) ∞
c) 0
d) -∞
View Answer

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.

Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
To practice all areas of Electromagnetic Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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