Magnetostatics MCQ Quiz - Objective Question with Answer for Magnetostatics - Download Free PDF

Explanation:

In a Magnetic Circuit:

Definition: A magnetic circuit is a path followed by magnetic flux. It is analogous to an electrical circuit but uses magnetic fields instead of electric currents. The primary components of a magnetic circuit include magnetic flux (Φ), magnetomotive force (MMF), and reluctance (R).

Working Principle: The magnetomotive force (MMF) in a magnetic circuit is created by a current passing through a coil of wire, creating a magnetic field. This MMF drives the magnetic flux through the magnetic circuit. The relationship between MMF, magnetic flux, and reluctance is given by:

MMF = Φ × R

where:

• MMF (magnetomotive force) is measured in Ampere-Turns (A-t).
• Φ (magnetic flux) is measured in Webers (Wb).
• R (reluctance) is measured in Ampere-Turns per Weber (A-t/Wb).

Reluctance: Reluctance is the opposition to the creation of magnetic flux in the magnetic circuit. It is analogous to resistance in an electrical circuit. The reluctance depends on the length (l) and cross-sectional area (A) of the magnetic path, as well as the material's permeability (μ), and is given by:

R = l / (μ × A)

where:

• l is the length of the magnetic path.
• μ is the permeability of the material.
• A is the cross-sectional area of the magnetic path.

Correct Option Analysis:

The correct option is:

Option 3: The magnetic flux will decrease.

Explanation: According to the relationship MMF = Φ × R, if the reluctance (R) increases and the magnetomotive force (MMF) remains constant, the magnetic flux (Φ) must decrease. This is because the reluctance provides greater opposition to the magnetic flux, thereby reducing its magnitude. An increase in reluctance means that it is harder for the magnetic flux to pass through the magnetic circuit, resulting in a decrease in the magnetic flux.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: The magnetomotive force (MMF) will increase.

This option is incorrect because the magnetomotive force (MMF) is a function of the current and the number of turns in the coil (MMF = N × I). An increase in reluctance does not inherently increase the MMF; instead, it affects the magnetic flux (Φ) for a given MMF.

Option 2: The magnetic flux will increase.

This option is incorrect because, as explained, an increase in reluctance leads to a decrease in magnetic flux, not an increase. The opposition to the magnetic flux becomes greater, reducing the amount of flux in the circuit.

Option 4: The resistance to magnetic flux will decrease.

This option is incorrect because reluctance is the magnetic equivalent of resistance. If the reluctance increases, the opposition (or resistance) to the magnetic flux increases, not decreases.

Conclusion:

Understanding the relationship between MMF, magnetic flux, and reluctance is crucial in analyzing magnetic circuits. When the reluctance of a path in a magnetic circuit increases, the magnetic flux decreases if the MMF remains constant. This fundamental principle helps in the design and analysis of magnetic circuits in various electrical and electronic applications.

The correct answer is: 4) The magnetic field lines spread out and become weaker.

Explanation:

• Magnetic fringing occurs at the ends of a magnetic circuit because:
• Magnetic field lines encounter lower reluctance in air compared to the core material, causing them to spread outward (fringe) rather than staying confined.
• This spreading leads to a weaker magnetic field near the edges, as some flux "leaks" instead of following the intended path.

Why Not the Other Options?

1. The MMF is not constant → MMF (magnetomotive force) depends on the current and turns; fringing occurs even with constant MMF.
2. The reluctance of the material increases → Reluctance is a property of the material and doesn’t inherently cause fringing.
3. The magnetic poles are not well-defined → While poles exist, fringing is due to flux spreading, not pole definition.

Explanation:

Magnetic Circuit and Reluctance

A magnetic circuit is a path followed by magnetic flux. It consists of materials with high magnetic permeability that guide the magnetic flux. Reluctance is the opposition that a magnetic circuit presents to the magnetic flux, analogous to electrical resistance in an electrical circuit. It is denoted by the symbol ℜ and is measured in Ampere-Turns per Weber (At/Wb).

Correct Option Explanation:

When the reluctance of one path in a series magnetic circuit increases, the total reluctance of the circuit also increases. According to the relationship Φ = MMF / ℜ, an increase in total reluctance (ℜ) results in a decrease in the total magnetic flux (Φ) if the MMF remains constant. This is because the magnetic flux is inversely proportional to the reluctance.

To understand this better, consider a series magnetic circuit with a constant MMF. If the reluctance of one section of the circuit increases, the opposition to the magnetic flux increases. As a result, the circuit's ability to carry magnetic flux diminishes, leading to a reduction in the total magnetic flux. This explains why the correct option is:

Option 4: The total flux will decrease.

 Additional Information

Let's analyze why the other options are incorrect:

• Option 1: The total flux will increase. This option is incorrect because an increase in reluctance means an increase in opposition to the magnetic flux. Therefore, the total flux cannot increase.
• Option 2: The total flux will remain unchanged. This option is incorrect because an increase in reluctance directly affects the total magnetic flux, causing it to decrease if the MMF is constant.
• Option 3: The flux will divide equally between paths. This option is incorrect because, in a series magnetic circuit, the flux does not divide between paths. The total flux is the same throughout the series circuit, and an increase in reluctance in one part affects the entire circuit's flux.

The correct understanding of magnetic circuits and reluctance is crucial for designing efficient magnetic systems. By ensuring that the reluctance is minimized, the total magnetic flux can be maximized, leading to better performance of magnetic devices such as transformers, inductors, and magnetic actuators.

Explanation:

The magnetic field outside a toroidal coil is zero. Let's delve into the detailed explanation and reasoning behind this conclusion.

Understanding the Toroidal Coil:

A toroidal coil, or simply a toroid, is a coil of wire in the shape of a doughnut or a closed loop. The wire is wound around a circular core, and the core itself can be either solid or hollow. The purpose of this design is to confine the magnetic field within the core and minimize the magnetic field outside the coil.

Magnetic Field in a Toroidal Coil:

To understand why the magnetic field outside a toroidal coil is zero, we need to consider the principles of electromagnetism, specifically Ampère's Law. Ampère's Law states that the line integral of the magnetic field B around any closed loop is proportional to the total current I passing through the loop. Mathematically, it is expressed as:

B × dl = μ₀ × I

where:

• B is the magnetic field
• dl is a differential element of the loop
• μ₀ is the permeability of free space
• I is the current passing through the loop

In the case of a toroidal coil, the wire is wound in such a way that the current flows in circular loops around the core. Because the coil is wound symmetrically and continuously around the core, the magnetic field inside the core is also circular and confined within the core. The magnetic field lines inside the toroid follow the loops of the wire, creating a strong and uniform magnetic field within the core.

Why the Magnetic Field Outside is Zero:

Outside the toroidal coil, the contributions of the magnetic field from each loop of the wire cancel each other out. This is due to the symmetry and closed-loop nature of the toroid. For any point outside the toroid, there are equal and opposite magnetic field contributions from different sections of the coil, resulting in a net magnetic field of zero.

Mathematically, this can be understood by applying Ampère's Law to a closed loop that lies outside the toroid. Since no net current passes through this loop (the current entering the loop at one point is exactly balanced by the current leaving the loop at another point), the line integral of the magnetic field around this loop is zero. Consequently, the magnetic field outside the toroid must be zero to satisfy Ampère's Law.

Therefore, the correct option is:

1) The magnetic field outside a toroidal coil is zero.

Concept:

In magnetic field theory, the work done on a unit North (N) pole in moving once around any closed path in a magnetic field is known as the magnetomotive force (MMF). According to Ampere’s Circuital Law, this MMF is equal to the total Ampere-turns enclosed by the path.

Ampere’s Law:

The line integral of magnetic field intensity \( H \) around a closed path is equal to the total current enclosed by the path:

\( \oint H \cdot dl = NI \)

Where \( NI \) is the total ampere-turns linked with the path.

Evaluation of Options:

Option 1: EMF in respective path –  Incorrect
EMF relates to electric circuits, not magnetic work done on poles.

Option 2: number of turns in respective path –  Incorrect
Turns alone don’t represent the work; current is also required.

Option 3: Ampere turns linked with the path –  Correct
This represents the correct physical quantity corresponding to the work done on a magnetic pole.

Option 4: current in respective path –  Incorrect
Current alone doesn't define the total magnetic work; number of turns must be considered too.

CONCEPT:

• Magnetic field strength or magnetic field induction or flux density of the magnetic field is equal to the force experienced by a unit positive charge moving with unit velocity in a direction perpendicular to the magnetic field.
  • The SI unit of the magnetic field (B) is weber/meter2 (Wbm-2) or tesla.
• The CGS unit of B is gauss.

​1 gauss = 10-4 tesla.

EXPLANATION:

• From the above explanation, we can see that the relation between tesla and Weber/m2 is given by:

1 tesla = 1 Weber/m2

CONCEPT:

• Fleming's Left-hand rule gives the force experienced by a charged particle moving in a magnetic field or a current-carrying wire placed in a magnetic field.
• It states that "stretch the thumb, the forefinger, and the central finger of the left hand so that they are mutually perpendicular to each other.
• If the forefinger points in the direction of the magnetic field, the central finger points in the direction of motion of charge, then the thumb points in the direction of force experienced by positively charged particles."

EXPLANATION:

• According to question

1. Forefinger (Index finger): Represents the direction of the magnetic field (magnetic flux). Therefore option 3 is correct.
2. Middle finger: Represents the direction of motion of charge (current).
3. The thumb : Represents the direction of force or motion experienced by positively charged particles.

Concept:

Magnetic Field Strength (H): the amount of magnetizing force required to create a certain field density in certain magnetic material per unit length.

\(H=\frac{NI}{L}\)

The intensity of Magnetization (I): It is induced pole strength developed per unit area inside the magnetic material.

The net Magnetic Field Density (Bnet) inside the magnetic material is due to:

• Internal factor (I)
• External factor (H)
   

∴ Bnet ∝  (H + I)

Bnet = μ0(H + I) …. (1)

Where μ0 is absolute permeability.

Note: More external factor (H) causes more internal factor (I).

∴ I ∝  H

I = KH …. (2)

And K is the susceptibility of magnetic material.

From equation (1) and equation (2):

Bnet = μ0(H + KH)

Bnet = μ0H(1 + K) …. (3)

Dividing equation (3) by H on both side

\(\frac{{{B_{net}}}}{H} = \frac{{{μ _0}H\left( {1 + K} \right)}}{H} \)

or, μ0μr = μ0(1 + K)

∴ μ_r = (1 + K) .... (4)

From equation (3) and (4)

Bnet = μ0μ_rH

Calculation:

Given Magnetic Circuit,

N = 100
I = 5 A
L = 2πr = 2π × 5 × 10^-2 m

From above concept,

\(H=\frac{NI}{L}=\frac{100× 5}{10π × 10^{-2}}=\frac{5000}{π}\)

We know that,

Bnet = μ0μrH

And, μ_r = 1000

\(B_{net}=4\pi \times 10^{-7}\times 1000\times \frac{5000}{\pi}=2\ T\)

Concept:

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

Radius (R) = 0.5 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 force between two current-carrying parallel conductors:

• Two current-carrying conductors attract each other when the current is in the same direction and repel each other when the currents are in the opposite direction
• Force per unit length on conductor

\(\frac{F}{l}=~\frac{{{\mu }_{0}}}{2\pi }\frac{{{i}_{1}}{{i}_{2}}}{d}\)

Faraday’s first law of electromagnetic induction states that whenever a conductor is placed in a varying magnetic field, emf is induced which is called induced emf. If the conductor circuit is closed, the current will also circulate through the circuit and this current is called induced current.

Faraday's second law of electromagnetic induction states that the magnitude of emf induced in the coil is equal to the rate of change of flux that linkages with the coil. The flux linkage of the coil is the product of number of turns in the coil and flux associated with the coil.

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)

μ₀ = Absolute permeability = 4π × 10^-7 H/m

Calculation:

Given:

r = 5 cm = 5 × 10^-2 m

I = 250 A

\(B = {\mu _0}H = {\mu _0}\frac{I}{{2\pi r}} = 4\pi \times {10^{ - 7}} \times \;\frac{{250}}{{2\pi\times 5 \times {{10}^{ - 2}}}} = {10^{ - 3}}\)

∴ B = 10^-3 Wb/m²

^{Important Points}

 Magnetic flux density of coil is given by ^{\(B = \;\frac{{{\mu _o}NI}}{{2R}}\)}  T

Magnetic flux density of solenoid is given by \(B = \;\frac{{{\mu _0}\;NI}}{l}\)  T

Magnetic flux density of long straight wire \(B = \;\frac{{{\mu _0}I}}{{2\pi r}}\) T

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)

μ0 = Absolute permeability = 4π × 10-7 H/m

Calculation:

Given:

r = 5 cm = 5 × 10-2 m

I = 250 A

\(H = \frac{I}{{2\pi r}} = \frac{{250}}{{2\pi\times 5 \times {{10}^{ - 2}}}} = \frac{2500}{\pi}\)

Important Points

 Magnetic flux density of coil is given by \(B = \;\frac{{{\mu _o}NI}}{{2R}}\)  T

Magnetic flux density of solenoid is given by \(B = \;\frac{{{\mu _0}\;NI}}{l}\)  T

Magnetic flux density of long straight wire \(B = \;\frac{{{\mu _0}I}}{{2\pi r}}\) T

Magnetic Field at the center of a Circular Current carrying coil:

Consider a current-carrying circular loop having its center at O carrying current i as shown below,

If dl is a small element at a distance r then, the magnetic field intensity on that point can be written using Biot-Savart’s law.

\(|\vec{dB}|=\frac{\mu_0i}{4π}\frac{idlsinθ}{r^2}\) .... (1)

Since, the loop is circular,

θ = 90° → sin θ = 1

Hence, equation (1) becomes,

\(|\vec{dB}|=\frac{\mu_0i}{4π}\frac{idl}{r^2}\)

If the circular loop is consisting of numbers of small element dl, then we will get the magnetic intensity all over the loop.

Hence to get the total field we must sum up that is integrate the magnetic field all over the field.

\(B=∫{\vec{dB}}\)

\(B=∫|\vec{dB}|=∫\frac{\mu_0i}{4π}\frac{idl}{r^2}\)

\(B=\frac{\mu_0i}{4π r^2}∫ dl\)

∫dl = l = 2πr

Hence,

\(B=\frac{\mu_oi}{2r}\)

Now, we have to find magnetic field intensity or magnetizing force (H),

And, H = \(\frac{B}{\mu_0}=\frac{i}{2r}\)

Calculation:

Given the length of the wire is 2 m and it bends to the circle, hence the circumference of the circle will be 2 m.

Now, We have, 2πr = 2 m, r = 1/π m
Magnetizing force at the centre of the circle, H = \(\frac{i}{2r}=\frac{50}{2\times1/π}=25π\)
H = 25π = 78.54 AT/m

CONCEPT: 

The magnetic field at point P due to a straight conductor is given by:

B = \(μ_0 I \over{2\pi d}\)

where B is the magnetic field at point P, μ0 permeability of the medium I is the current in the in the wire and d is the distance from the wire to that point.

EXPLANATION:

At distance r from an infinitely long straight current carrying conductor, magnetic field is:

B = \(μ_0 I \over{2\pi r}\)

where B is the magnetic field, μ0 permeability of the medium I is the current in the wire, and r is the distance from the wire to that point.

So B α 1/r

Important Points

 Magnetic flux density of coil is given by \(B = \;\frac{{{\mu _o}NI}}{{2R}}\)  T

Magnetic flux density of solenoid is given by \(B = \;\frac{{{\mu _0}\;NI}}{l}\)  T

Magnetic flux density of long straight wire \(B = \;\frac{{{\mu _0}I}}{{2\pi r}}\) T