# Magnetostatics MCQ || Magnetostatics Questions and Answers

Ques.1. Find H = ___________ A/m at the center of a circular coil of diameter 1 m and carrying a current of 2 A.

1. 0.6366
2. 0.1636
3. 6.366
4. 2

Explanation:-

The magnetic field intensity (H) of a circular coil is given by

$H = \frac{I}{{2 R}}$

Where I is the current flow through the coil

R is the radius of the circular coil

Calculation:

Given that, Current (I) = 2 A

Diameter = 1 m

Magnetic field intensity $H = \frac{2}{{2 \times 0.5}} = 2\;A/m$

Common Mistake:

The magnetic field intensity (H) of a circular coil is given by $H = \frac{I}{{2 R}}$

The magnetic field intensity (H) of a straight conductor is given by $H = \frac{I}{{2\pi R}}$

Ques.2. Hysteresis loss is ________ proportional to the area under the hysteresis curve. Also, it is ________ proportional to the number of cycles of magnetization per second:

1. Directly, inversely
2. Inversely, directly
3. Directly, directly
4. Inversely, inversely

Explanation:-

The hysteresis loss is directly proportional to the area under the hysteresis curve i.e. area of the hysteresis loop. It is directly proportional to frequency i.e. number of cycles of magnetization per second.

Hysteresis loss occurring in a material is,

Wh = η × Bm1.6 × f × V

Where η is hysteresis constant

f is frequency or number of cycles per second

Bm is the magnetic flux density

V is the volume of the core

Hence it is directly proportional to the number of cycles of magnetization per second

Ques.3. Which of the following materials is used for the generation of ultrasonic waves by using magnetostriction effect?

1. Paramagnetic material
2. Ferromagnetic material
3. Diamagnetic materials
4. Both paramagnetic and diamagnetic material

Explanation:-

Ferromagnetic substance or material is used for the generation of ultrasonic waves by using the magnetostriction effect.

Magnetostriction Effect:

• When a magnetic field is applied parallel to the length of a ferromagnetic rod made of a material such as iron or nickel, a small elongation or contraction occurs in its length. This is known as magnetostriction.
• The change in length depends on the intensity of the applied magnetic field and the nature of the ferromagnetic material.
• The change in length is independent of the direction of the field.

Ques.4. If a circular conductor carries a current of ‘I’ ampere having radius ‘r’ meter, then the magnetizing force at the center of the coil is given by ;

1. I/4r AT/m
2. I/r AT/m
3. I/2r AT/m
4. I/2r AT/wb

Explanation:-

Magnetic Field Strength (H)  gives the quantitative measure of the strongness or weakness of the magnetic field.

H = B/μo

Where

B = Magnetic Flux Density

μ= Vacuum Permeability

The magnetic Field strength at the center of circular loop carrying current I is given by

B = μoI/2r

B/μo = I/2r

H = I/2r At/m

Ques.5. What is the relationship between magnetic field strength and current density?

1. ∇.H = J
2. ∇.J = H
3. ∇ × H = J
4. ∇ × J = H

Answer.3. ∇ × H = J

Explanation:-

1) Modified Kirchhoff’s Current Law:

$\nabla .\vec J + \frac{{\partial \rho }}{{\partial t}} = 0$

J = Conduction Current density

2) Modified Ampere’s Law:

$\nabla \times \vec H = \vec J + \frac{{\partial \vec D}}{{\partial t}}$

Where $\frac{{\partial \vec D}}{{\partial t}}$ = Displacement current density

$\nabla .\vec E = – \frac{{\partial \vec B}}{{\partial t}}$

4) Gauss Law:

$\nabla .\vec D = \rho$

Maxwell’s Equations for time-varying fields is as shown:

 Differential form Integral form Name $\nabla \times E = – \frac{{\partial B}}{{\partial t}}$ $\mathop \oint \nolimits_L^{} E.dl = – \frac{\partial }{{\partial t}}\mathop \smallint \nolimits_S^{} B.d S$ Faraday’s law of electromagnetic induction $\nabla \times H =J+ \frac{{\partial D}}{{\partial t}}$ $\mathop \oint \nolimits_L^{} H.dl = \mathop \smallint \nolimits_S^{} (J+\frac{{\partial D}}{{\partial t}}).dS$ Ampere’s circuital law ∇ . D = ρv $\mathop \oint \nolimits_S^{} D.dS = \mathop \smallint \nolimits_v^{} \rho_v.dV$ Gauss’ law ∇ . B = 0 $\mathop \oint \nolimits_S^{} B.dS = 0$ Gauss’ law of Magnetostatics (non-existence of magnetic monopole)

Ques.6. Two identical coils A and B of 1000 turns each lie in the parallel plane such that 80% of the flux produced by one coil links with the other. If a current of 5 A flowing in A produces a flux of 0.05 mWb, then the flux linking with coil B is:

1. 0.4 mWb
2. 0.04 mWb
3. 4 mWb
4. 0.004 mWb

Explanation:-

Consider two coils having self-inductance L1 and L2 placed very close to each other. Let the number of turns of the two coils be N1 and N2 respectively. Let coil A carries current I1 and coil B carries current I2.

Due to current I1, the flux produced is ϕ1 which links with both the coils. Then the mutual inductance between two coils can be written as

$M = \frac{{{N_1}{\phi _{12}}}}{{{I_1}}}$

Here, ϕ12 is the part of the flux ϕ1 linking with the coil 2

Calculation:

Flux produced in coil X (ϕ1) = 0.05 mWb

As we are just required to find the flux linked with the second coil, and we are given that 80% of the flux produced by one coil links with the other.

∴ Flux linked with Y (ϕ12) = 80% of flux produced in coil 1

= 0.05 × 0.8 mWb

0.04 mWb

Ques.7. The current flowing through a coil of 2 H inductance is decreasing at a rate of 4 A/s. What will be the induced EMF in the coil?

1. 8 V
2. – 8 V
3. – 4 V
4. 4 V

Explanation:-

Emf induced in a coil or inductor is given by

$E = – N\frac{{d\phi }}{{dt}} = – L\frac{{di}}{{dt}}$

Where,

L = Inductance

di = final value of current – initial value of current

dt = final time – initial time

Given:

L = 2 H, di/dt = – 4 A/s (due to decreasing rate)

Therefore,

$E = – L{\text{\;}}\frac{{di}}{{dt}}$

= – (2 × -4)

8 V

Ques.8. Magnetic flux will be _________ if the surface area vector of a surface is perpendicular to the magnetic field.

1. Zero
2. Unity
3. Close to maximum
4. Maximum

Explanation:-

The magnetic flux is defined as the number of magnetic field lines passing through a closed surface

Flux can be expressed as

φ = BACosθ

ϕ  is the magnetic flux

B is the magnetic flux density

A is the area

θ is the angle between the surface area vector of a surface and magnetic field

Given that, θ = 90°

φ = BACos90° = 0

Ques.9. The force on the current-carrying conductor in a magnetic field depends upon:

(a) the flux density of the field

(b) The strength of the current

(c) The length of the conductor perpendicular to the magnetic field

(d) The directions of the field and the current

1. (a), (c) and (d) only
2. (a), (b)and (c) only
3. (a), (b) and (d) only
4. (a), (b), (c) and (d)

Answer.4. (a), (b), (c) and (d)

Explanation:-

The force experienced by a current-carrying conductor lying in a magnetic field is with an angle θ is given by

F = BIL sin θ

Where,

B = Magnetic flux density of the field

I = Current

L = Length of the conductor perpendicular to the magnetic field

Here angle θ represents the direction of the field and the current.

Therefore, the force on the current-carrying conductor in a magnetic field depends upon

• The magnetic flux density of the field (B)
• The strength of the current (I)
• The length of the conductor perpendicular to the magnetic field (l)
• The directions of the field and the current

Ques.10. Calculate the flux density at a distance of 5 cm from a long straight circular conductor carrying a current of 250 A and placed in air.

1. 102 Wb/m2
2. 10-2 Wb/m2
3. 10-3 Wb/m2
4. 103 Wb/m2

Explanation:-

The magnetizing field strength due to long straight circular conductor is given by

$H = \:\frac{I}{{2\pi r}}\:AT/m$

Where, H = Magnetizing force (AT/m)

I = Current flowing in a conductor (A)

r = Distance between current carrying conductor and the point (m)

also B = μ0 H

Where, B = Magnetic flux density (Wb/m2)

μ= Absolute permeability = 4π × 10-7 H/m

Given:

r = 5 cm = 5 × 10-2 m

I = 250 A

$\begin{gathered} B = {\mu _0}H \hfill \\ \hfill \\ = {\mu _0}\frac{I}{{2\pi r}} \hfill \\ \hfill \\ = 4\pi \times {10^{ – 7}} \times \:\frac{{250}}{{2\pi \times 5 \times {{10}^{ – 2}}}} = {10^{ – 3}} \hfill \\ \end{gathered}$

∴ B = 10-3 Wb/m2

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