Magnetic Flux Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. What is the SI unit of magnetic flux?

1. Tesla
2. Weber
3. Dyne
4. Gauss

Ans. The correct answer is b. Weber

Explanation: The SI unit of magnetic flux is Weber (Wb).

Ques. The ratio of magnetic force to electric force on a charged particle getting undeflected in a field is?

1. 0
2. 2
3. 1
4. 4

Ans. The correct answer is c. 1

Explanation: Since the strength of the magnetic and electric forces acting on a charged particle in a field are equal when the particle is undeflected, the ratio equals 1.

Ques. What is the dimensional formula of magnetic flux?

1. [M L² T^-2 A^-1]
2. [M² L² T^-3 A^-1]
3. [M L T^-2 A^-2]
4. [M L3 T^-3 A^-1]

Ans. The correct answer is a. [M L² T^-2 A^-1]

Explanation: The dimensional formula of magnetic flux is [M L² T^-2 A^-1].

Ques. What is the dimensional formula of magnetic flux density?

1. [M² T^-2 A^-1]
2. [M T^-3 A^-1]
3. [M T^-2 A^-1]
4. [M T^-2 A^-2]

Ans. The correct answer is c.[M T^-2 A^-1]

Explanation: The dimensional formula of magnetic flux density is [M T^-2 A^-1]

Ques. The CGS unit of magnetic flux is

1. Emu
2. Maxwell
3. Erg
4. Dyne

Ans. The correct answer is b. Maxwell

Explanation: The CGS unit of magnetic flux is Maxwell (Mx)

Ques. The CGS unit of magnetic flux density is

1. Erg
2. Maxwell
3. Dyne
4. Gauss

Ans. The correct answer is d. Gauss

Explanation: The CGS unit of magnetic flux density is Gauss

Ques. When a charged particle moves at right angles to the magnetic field, the variable quantity is?

1. Speed
2. Momentum
3. Moment of inertia
4. Energy

Ans. The correct answer is b. Momentum

Explanation: When a charged particle moves perpendicular to the field, its speed remains constant while its velocity changes. Momentum is the product of the particle's mass and velocity; therefore, when velocity changes, so does momentum.

Ques. The magnetic field outside an ideal solenoid is

1. Infinity
2. Negative
3. Zero
4. One

Ans. The correct answer is c. Zero

Explanation: The magnetic field outside an ideal solenoid is zero.

Ques. Which, among the following qualities, is not affected by the magnetic field?

1. Current flowing in a conductor
2. Stationary charge
3. Change in magnetic flux
4. Moving charge

Ans. The correct answer is b. Stationary charge

Explanation: Because stationary charges don't possess any motion, they are unaffected by magnetic fields. A particle with no velocity cannot have a magnetic field.

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Short Answers Questions [2 Marks Questions]

Ques. Define magnetic flux.

Ans. The number of magnetic field lines traveling through a given closed surface is defined as magnetic flux. It measures the total magnetic field that travels across a specific surface area. In this case, the area under consideration can be of any size and oriented in any direction in relation to the magnetic field's direction.

Ques. What is the formula of magnetic flux?

Ans. Magnetic flux can be given by the product of the magnitude of the magnetic field passing through a surface at an angle θ and the area of that surface.

Φ = BA cosθ

Where

• Φ is the magnetic flux
• B is the strength of the magnetic field
• A is the area of the surface
• θ is the angle between the magnetic field and the area of the surface.

Ques. Define magnetic flux density.

Ans. Magnetic flux density (B) is defined as the force acting on a wire at right angles to the magnetic field per unit current per unit length.

It is a vector quantity and its SI unit is Tesla (T).

Ques. What is the condition for maximum and minimum magnetic flux?

Ans. Magnetic flux is given by the formula

Φ = BA cosθ

Magnetic flux will be maximum if θ = 0 i.e. the magnetic field is parallel to the area vector or perpendicular to the plane of the area.

Φ_max = BA cos 0 = BA

Magnetic flux will be minimum if θ = 90 i.e. the magnetic field is perpendicular to the area vector or parallel to the plane of the area.

Φ_max = BA cos 90 = 0

Ques. Define magnetic susceptibility.

Ans. Magnetic susceptibility is the property of the substance which shows how easily the substance can be magnetized when placed in the magnetizing field. It is denoted by (χ_m). It has no unit. It is just a number.

Ques. Define magnetic permeability.

Ans. The extent to which magnetic field lines can enter a substance is known as magnetic permeability. It is denoted by µ.

It is given by the ratio of the strength of the magnetic field to the intensity of the magnetizing field.

Ques. Define relative magnetic permeability.

Ans. Relative magnetic permeability is defined as the ratio of the flux density inside the material to the flux density in a vacuum. It is dimensionless quantity.

Also Read:

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Long Answers Questions [3 Marks Questions]

Ques. A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. What is the coercivity of the bar magnet?

Ans. Given

• Length of the solenoid, L = 0.2 m
• Number of turns of the solenoid, N = 100 turns
• Current flowing through the solenoid, I = 5.2 A

The coercivity of the bar magnet is given by

H = B/µ₀ 

But for a solenoid, B = µ₀nI

Therefore, coercivity, H = µ₀nI/µ₀ = nI

Where n is the number of turns per unit length of the solenoid.

⇒ H = (N/L)I

On substituting the values, we get

H = (100/0.2) x 5.2

⇒ H = 2600 A/m

Ques. A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20 x 10^-6 J/T when a magnetic intensity of 60 x 10³ A/m is applied. What is its magnetic susceptibility?

Ans. Given

• Length of each side of cubic paramagnetic substance, L = 1 cm = 10^-2 m
• Magnetic dipole moment, m = 20 x 10^-6 J/T
• Magnetic intensity, H = 60 x 10³ A/m

The magnetic susceptibility of the paramagnetic substance is given by

χ_m = Intensity of magnetization(I)/Magnetic intensity(H)

But Intensity of magnetization, I = Magnetic moment(m)/Volume(V)

⇒ χ_m = m/HV

Volume of the given substance, V = L3 = 10^-6 m³

On substituting the values, we get

χ_m = (20 x 10^-6)/(60 x 10³ x 1^0-6)

⇒ χ_m = 3.3 x 10^-4

Ques. What is the difference between electric flux and magnetic flux?

Ans. The following are the differences between electric flux and magnetic flux

Ques. What are the applications of magnetic flux?

Ans. The following are the applications of magnetic flux

• Generators: Generators use magnetic flux to transform mechanical energy into electrical energy. They are utilized for many different purposes, including power generation and wind turbines.
• Electric motors: Magnetic flux is used by electric motors to transform electrical energy into mechanical energy. They can be used in many different applications, including appliances, electrical devices, and transportation.
• Loudspeakers: Magnetic flux is used in loudspeakers to change electrical energy into sound waves. They are utilized in a wide range of applications, including home entertainment and public speech systems.
• Magnetic resonance imaging (MRI): Strong magnetic fields are used by MRI equipment to generate images of the interior of the human body.

Ques. What are the properties of magnetic flux?

Ans. The following are the properties of magnetic flux

• The scalar product (dot product) of the magnetic field vector (B) and the elementary area vector (dA) is magnetic flux. As a result, it is a scalar quantity.
• The SI unit of magnetic flux is Weber (Wb). One Weber is equivalent to one Volt-Second.
• Moving electric charges or changing electric fields produce magnetic flux, as defined by Faraday's law of electromagnetic induction.
• Magnetic flux is conserved in nature. This indicates that the total magnetic flux through any closed surface is always zero.

Ques. What are the different ways to increase the magnetic flux through a coil?

Ans. The magnetic flux through a coil can be increased by

• Increasing the number of turns in the coil: If the number of turns of wire in the coil is more, the stronger the magnetic field.
• Increasing the current passing through the coil: The more current that flows through a coil, the stronger the magnetic field it produces.
• Making use of a ferromagnetic core: Magnetic fields attract and concentrate ferromagnetic elements such as iron and nickel. By inserting a ferromagnetic core into a coil, the magnetic flux across the coil is increased.
• Decreasing the area of the cross-section of the coil: The smaller the cross-sectional area of a coil, the more strong the magnetic field. This is due to the fact that the same amount of magnetic flux lines will pass through a smaller region.

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Very Long Answers Questions [5 Marks Questions]

Ques. A long solenoid having 5000 turns/m carries a current of 2.0 A. Calculate

1. Magnetic intensity
2. Magnetic flux density at the center of the solenoid

Now, an iron rod is inserted is inserted in the solenoid. Calculate

1. Magnetic intensity
2. Magnetic flux density at the center of the solenoid

The magnetic susceptibility of iron is 5 x 10³

Ans. Given

• Number of turns per unit length of the solenoid, n = 5000 turns/m
• Current flowing through the solenoid, I = 2.0 A
• The magnetic susceptibility of iron, χ_m = 5 x 10³

First case: In the absence of an iron rod

1. Magnetic intensity is given by

H = nI

⇒ H = 5000 x 2 = 10000 A m^-1

2. Magnetic flux density at the center of the solenoid is given by

B₀ = µ₀H

Where µ₀ is the absolute permeability

⇒ B₀ = 4π x 10^-7 x 10000

⇒ B₀ = 12.57 x 10^-3 T

Second case: In the presence of an iron rod

1. Magnetic intensity will remain the same i.e.

H = nI = 5000 x 2 = 10000 A m^-1

2. Magnetic flux density at the center of the solenoid when an iron rod is inserted in the solenoid is given by

B = µ₀ (H + M) …(i)

Where

• H is the magnetic intensity
• M is the intensity of magnetization

The intensity of magnetization is given by

M = χ_mH

⇒ M = 5 x 10³ x 10000 = 5 x 10⁷ A m^-1

On substituting the values in equation (i), we get

B = 4π x 10^-7 (10⁴ + 5 x 10⁷)

⇒ B = 62.85 T

Ques. Calculate the permeability and susceptibility of a magnetic bar of cross-section 0.1 cm² having a magnetic flux of 2.41 x 10^-5 weber due to magnetic intensity of 3200 A m^-1.

Ans. Given

• Magnetic intensity, H = 3200 A m^-1
• Magnetic flux, Φ = 2.41 x 10^-5 weber
• Area of the cross-section, A = 0.1 cm² = 0.1 x 10^-4 m²

The permeability of the magnetic bar is given by the ratio of the magnetic field strength to magnetic intensity i.e.

µ = B/H

But magnetic field strength, B = Φ/A

⇒ µ = Φ/AH

On substituting the values, we get

µ = (2.41 x 10^-5 )/(0.1 x 10^-4 x 3200)

⇒ µ = 7.53 x 10^-4 T m A^-1

Also, the permeability of the magnetic bar is given by

µ = µ₀ (1 + χ_m)

⇒ χ_m = (µ/µ₀) - 1

Where

• χ_m is the magnetic susceptibility
• µ₀ is the absolute permeability

On substituting the values, we get

χ_m = [(7.53 x 10^-4 )/(4π x 10^-7)] - 1

⇒ χ_m = 596.1

Ques. An ideal solenoid having 2000 turns per meter has an iron core of relative permeability of 500 and carries a current of 1.0 A. Calculate

1. Magnetic intensity (H) at the center of the solenoid
2. The magnetic permeability of the iron
3. Magnetic susceptibility of iron
4. Magnetic field B
5. Magnetization (M_net)
6. Magnetising current (I_m)

Ans. Given

• The number of turns per unit length of the solenoid, n = 2000 turns/m
• Relative permeability of the iron core, µr = 500
• The current flowing through the solenoid, I = 1.0 A

1. Magnetic intensity at the center of the solenoid is given by

H = nI = 200 x 1 = 2000 A m^-1

2. The magnetic permeability of the iron is given by

µ = µ₀µr

⇒ µ = 4π x 10^-7 x 500 = 6.28 x 10^-4 T m A^-1

100. The magnetic susceptibility of iron is given by

χ_m = µ_r - 1 = 500 - 1 = 499

500. Magnetic field, B = µH

⇒ B = 6.28 x 10^-4 x 2000 = 1.26 T

5. Magnetization, M_net = χ_mH

⇒ M_net = 499 x 2000 = 9.98 x 10⁵ A m^-1

6. Magnetizing current is the current required to get the same value of flux density B inside the solenoid as available when a magnetic core is used in the solenoid. Magnetizing current is given by

I_m = (B/µ₀n) - I

⇒ I_m = (1.26/4π x 10^-7 x 2000) - 1 = 5000.6 A

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