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

Composite Magnetic Circuit with Two Magnetic Materials

Introduction: In a composite magnetic circuit, the circuit consists of different magnetic materials with varying permeabilities (μ1 and μ2). Each material in the circuit contributes to the total reluctance of the system. Reluctance is the magnetic equivalent of electrical resistance and is inversely proportional to the permeability of the material.

Total Reluctance in a Composite Magnetic Circuit:

The total reluctance of a composite magnetic circuit is analogous to the total resistance in an electrical circuit. When magnetic materials with different permeabilities are combined, the overall reluctance of the circuit is determined by summing up the reluctances of each material. This is because the magnetic flux travels through each material sequentially, similar to a series electrical circuit. Mathematically, the total reluctance R_total in a composite magnetic circuit is given by:

R_total = R₁ + R₂ + ... + R_n

Where:

• R_i represents the reluctance of the i-th material.

The reluctance of a magnetic material is calculated using the formula:

R = l / (μ * A)

Where:

• R = Reluctance of the material (measured in Ampere-Turns per Weber, A·T/Wb)
• l = Length of the magnetic path through the material (in meters)
• μ = Permeability of the material (in Henrys per meter, H/m)
• A = Cross-sectional area of the material (in square meters)

Since reluctance is inversely proportional to permeability, a material with higher permeability will have lower reluctance, and vice versa. In a composite magnetic circuit, the reluctances of all the materials are added together to determine the total reluctance of the circuit.

Correct Option Analysis:

The correct option is:

Option 3: The sum of reluctances of each material.

This option correctly describes the calculation of total reluctance in a composite magnetic circuit. When multiple materials with different permeabilities are present in the circuit, their reluctances are added together to determine the total reluctance of the circuit. The summation approach is justified because the magnetic flux travels sequentially through each material, and the reluctances act like resistances in a series electrical circuit.

Explanation:

Artificial magnets are man-made magnets, specifically designed and manufactured to possess desired magnetic properties. These magnets are utilized in various applications, including electrical motors, transformers, sensors, and other devices that rely on magnetic fields for operation. Their efficiency, stability, and durability are critical factors in determining their suitability for specific applications.

Correct Option Analysis:

The correct option is:

Option 1: Ceramic Magnet

Ceramic magnets, also known as ferrite magnets, are the best-suited artificial magnets for use in electrical motors and transformers due to their high efficiency and stable magnetic field. These magnets are composed of a ceramic material that includes iron oxide combined with other elements such as barium or strontium. They are widely recognized for their excellent magnetic properties, cost-effectiveness, and versatility in various applications.

Key Features of Ceramic Magnets:

• High Stability: Ceramic magnets exhibit a stable magnetic field over time, making them ideal for applications where consistency is crucial, such as in electrical motors and transformers.
• Resistant to Demagnetization: These magnets are highly resistant to demagnetization, ensuring long-term reliability in demanding environments.
• Cost-Effective: Ceramic magnets are relatively inexpensive to produce, making them an economical choice for industrial applications.
• Wide Operating Temperature Range: They can operate effectively over a broad temperature range without significant loss of magnetic properties.
• Corrosion Resistance: Ceramic magnets are resistant to corrosion, allowing them to function efficiently in various environmental conditions.

The correct answer is  3) Soft magnetic material

Explanation:

• A soft magnetic material is easily magnetized when an external magnetic field is applied and loses its magnetism when the field is removed.

• It has low coercivity and low retentivity, making it ideal for applications like transformer cores and electromagnets where temporary magnetism is needed.

Additional Information

• 1) Hard magnetic material:
  Retains magnetism even after the external field is removed (used in permanent magnets).

• 2) Ferromagnetic material:
  A general term for materials (like iron, cobalt, nickel) that show strong magnetic properties, which can be soft or hard.

• 4) Permanent magnet:
  Made of hard magnetic material; keeps its magnetism permanently.

High Permeability in Magnetic Materials

Definition: Permeability is a property of a magnetic material that measures its ability to support the formation of a magnetic field within itself. High permeability indicates that the material allows magnetic field lines to pass through it more easily, making it an ideal choice for use in magnetic circuits.

Magnetic Circuit and Permeability:

In a magnetic circuit, magnetic flux is analogous to current in an electrical circuit, and reluctance is analogous to resistance. The relationship between these parameters can be expressed by the magnetic version of Ohm's Law:

Φ = MMF / Reluctance

Where:

• Φ is the magnetic flux
• MMF is the magnetomotive force
• Reluctance (R_m) is the opposition to the formation of magnetic flux, analogous to electrical resistance

The reluctance of a magnetic circuit is inversely proportional to the permeability of the material. That is:

R_m = l / (μ × A)

Where:

• l = Length of the magnetic path
• μ = Permeability of the material
• A = Cross-sectional area of the material

From this relationship, it is clear that as the permeability (μ) of a material increases, the reluctance (R_m) decreases. This means that the material offers less opposition to the magnetic flux, making it easier for the flux to pass through the material.

Correct Option Analysis:

The correct option is:

Option 1: Decreased reluctance

When a magnetic material has high permeability, its reluctance decreases. This is because reluctance is inversely proportional to permeability, as shown in the formula above. A material with high permeability allows magnetic flux to pass through it more readily, reducing the overall reluctance of the magnetic circuit. This property is particularly important in designing magnetic cores for transformers, inductors, and other electromagnetic devices, where minimizing reluctance is essential for efficient operation.

Explanation:

Magnetic Flux and Reluctance

Definition: Magnetic flux (Φ) is a measure of the total magnetic field passing through a given area. It is expressed in Weber (Wb). Reluctance (ℜ), on the other hand, is the opposition offered by a magnetic circuit to the flow of magnetic flux, analogous to resistance in an electrical circuit. Reluctance is given by the formula:

ℜ = l / (μ × A)

Where:

• l = Length of the magnetic path (in meters)
• μ = Permeability of the material (in Henry/meter)
• A = Cross-sectional area of the magnetic path (in square meters)

The relationship between the magnetomotive force (MMF), reluctance, and magnetic flux is given by:

Φ = MMF / ℜ

Where:

• Φ = Magnetic flux (in Weber)
• MMF = Magnetomotive force (in Ampere-turns)
• ℜ = Reluctance (in Ampere-turns per Weber)

Correct Option Analysis:

The correct option is:

Option 4: The flux will decrease.

If the length of the magnetic path is increased while keeping the cross-sectional area (A) and magnetomotive force (MMF) constant, the reluctance of the magnetic circuit will increase. This is because reluctance (ℜ) is directly proportional to the length of the magnetic path (l). As reluctance increases, the magnetic flux (Φ), which is inversely proportional to reluctance, will decrease. This can be mathematically expressed as:

Φ = MMF / ℜ

With a higher reluctance (ℜ) due to an increased length (l) of the magnetic path, the denominator in the above equation becomes larger, resulting in a smaller value of Φ. Thus, the magnetic flux will decrease

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 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:

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