Application of Ampere's Circuit Law MCQ Quiz in मल्याळम - Objective Question with Answer for Application of Ampere's Circuit Law - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT:

Magnetic Induction in a Solenoid

• The magnetic field (B) at the center of a long solenoid is given by the formula:

  B = μ₀ × n × I

• Where:
  • B = Magnetic induction (T)
  • μ₀ = Permeability of free space = 4π × 10⁻⁷ H/m
  • n = Number of turns per meter = N / L
  • I = Current (A)

EXPLANATION:

• Given:
  • N = 2000 turns
  • L = 0.5 m
  • B = 0.08 T

n = N / L = 2000 / 0.5 = 4000 turns/m

0.08 = (4π × 10⁻⁷) × 4000 × I

I = 0.08 / [(4π × 10⁻⁷) × 4000]

I = 0.08 / (16π × 10⁻⁴)

I ≈ 0.08 / (0.005027) ≈ 15.91 A ≈ 16 A

Therefore, the correct answer is 16 A

Concept:

Magnetization in a Solenoid:

• Magnetization (M) is the measure of the magnetic moment per unit volume in a material. It is related to the magnetic field intensity (H) and the relative permeability of the material (μr).
• For a solenoid, the magnetic field intensity H is given by the formula:
  • H = nI, where n is the number of turns per unit length (500 turns/m), and I is the current (3 A).
• The magnetization M is related to H and the relative permeability μr as:
  • M = (μr - 1) H, where μr is the relative permeability of the material (5001 in this case).

Calculation:

Given,

Number of turns per unit length: n = 500 turns/m

Current through the solenoid: I = 3 A

Relative permeability of the core: μr = 5001So, O

The magnetic field intensity H is:

H = nI

H = 500 × 3 = 1500 A/m

Now, using the formula for magnetization M:

M = (μr - 1) H

M = (5001 - 1) × 1500

M = 5000 × 1500 = 7.5 × 10⁶ A/m

∴ The magnitude of magnetization is 7.5 × 10⁶ A/m. So, option 3) is correct.

Maxwell Equations:

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

3) Faraday’s Law:

\(\nabla .\vec E = - \frac{{\partial \vec B}}{{\partial t}}\)

4) Gauss Law:

\(\nabla .\vec D = \rho \)

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

Solution:

Magnetic field (B) generated by one sheet at the location of the second is given by:
B = (μ₀ K)/2 (direction parallel to the second sheet)

Force (F) acting on a sheet of width 'b' and length 'ℓ' is:
F = (b × K × ℓ) × (μ₀ K/2)

Force per unit area (P) is then:
P = F / (ℓ × b)
P = (μ₀ K²)/2

By substituting the given current per unit width (K = β / 2√π), we have:

P = (μ₀ / 2) × (β² / 4π)

⇒ P = (4π × 10⁻⁷) × (β² /8π) =2 × 10⁻⁷ N/m²

⇒ β = 2

Concept:

Magnetic Field Inside a Solenoid with a Core Material:

• The magnetic field inside a solenoid with a core of relative permeability (μ_r) is given by:
• B = μ₀ μ_r n I
  where: μ₀ = permeability of free space (4π × 10^-7 T·m/A), μ_r = relative permeability of the core material, n = number of turns per unit length, I = current.

Calculation:

Given,

Relative permeability, μ_r = 400

Current, I = 1 A

Number of turns per metre, n = 1000 turns/m

μ₀ = 4π × 10^-7 T·m/A

Now, calculate the magnetic field B:

B = μ₀ μ_r n I

B = (4π × 10^-7) × 400 × 1000 × 1

B = 16π × 10^-2 T

∴ The magnetic field is 16π × 10^-2 T.
Hence, the correct option is 3)

Concept:

Magnetic Field on the Axis of a Circular Loop:

• The magnetic field at a point along the axis of a circular current loop is given by B = (μ₀ I R²) / (2 (R² + x²)^3/2), where x is the distance from the center of the loop.
• We are given the ratio of the magnetic fields at two points A and B and need to find the radius R of the loop.

Calculation:

Magnetic field at A: B₁ = (μ₀ I R²) / (2 (R² + 4²)^3/2)

Magnetic field at B: B₂ = (μ₀ I R²) / (2 (R² + (3√3)²)^3/2)

Given: B₁ / B₂ = 216 / 125

After simplifying the ratio:

(R² + 27)^3/2 / (R² + 16)^3/2 = 1.728

Taking cube roots:

(R² + 27)^1/2 / (R² + 16)^1/2 = 1.2

(R² + 27) / (R² + 16) = 1.44

Solving:

R² ≈ 9

R ≈ 3 cm

∴ The radius of the loop is 3 cm. Option 1) is correct.

\(\text{Solution: Magnetic field inside the wire is } B_r = \frac{\mu_0 I r}{2\pi a^2} \Rightarrow B_p = \frac{\mu_0 I (a/3)}{2\pi a^2} \\ \text{Magnetic field outside the wire is } B = \frac{\mu_0 I}{2\pi r} \Rightarrow B_o = \frac{\mu_0 I}{2\pi (9a)} \\ \text{Thus } \frac{B_p}{B_o} = \frac{\frac{\mu_0 I (a/3)}{2\pi a^2}}{\frac{\mu_0 I}{2\pi (9a)}} = 3\)

The correct answer is 3.

Concept:

Calculation: 

µ₀ni = B n = number of turns per unit length 

\(\mu_0\left(\frac{\mathrm{m}}{\ell}\right) \mathrm{i}=\mathrm{B}\)

\(\mathrm{m}=\frac{\mathrm{B} \cdot \ell}{\mu_0 \mathrm{i}}=\frac{6.28 \times 10^{-3} \times 0.5}{12.56 \times 10^{-7} \times 5}\)

m = 500  

∴ The value of m is 500.

Concept:

Magnetic Force: The magnetic force on a current-carrying conductor is given by:

\( F_{\text{magnetic}} = I \times L \times B \)

Where, I is Current through the wire, L is Length of the wire, B is Magnetic field strength

Gravitational Force: The gravitational force on the wire is given by:

\( F_{\text{gravity}} = m \times g \)

For the wire to be suspended in mid-air

\( F_{\text{magnetic}} = F_{\text{gravity}} \)

The magnetic field  B  can be calculated as

\( B = \frac{m \times g}{I \times L} \)

Calculation:

Here,

I = 2 A, R = 1 Ω, ρ = 2 × 10^-6 Ωm, A = 10 mm² = 10 × 10^-6 m²

m = 500 g = 0.5 kg, g = 10 m/s²

Length of the Wire (L),  \( L = \frac{R \times A}{\rho} = \frac{1 \times 10 \times 10^{-6}}{2 \times 10^{-6}} = 5 \, \text{m} \)

Calculate the Magnetic Field (B), \( B = \frac{m \times g}{I \times L} = \frac{0.5 \times 10}{2 \times 5} = 0.5 \, \text{T} = 5 \times 10^{-1} \, \text{T} \)

∴ The magnitude of the magnetic field B is 5 × 10^-1 T.

Solenoid:

• Solenoid is a type of electromagnet which is a coils of wire.

• Its purpose is to generate a magnetic field which is controlled by some value of current through a coil which is wound into a tightly packed helix.
• The coil can be arranged to produce a uniform magnetic field in a vacuum of space when an electric current is passed through it.

In the question given that there are two solenoids.

- One is made of thick wire

- Other is made of thin wire

⇒ Area of thick wire (A₁) > area of thin wire (A₂)

Now parameters for comparison of the solenoids are

i) Self inductance (L):

• \(L = \frac{\phi }{I} = \frac{{\mu {N^2}\left( {Hollow\;area\;of\;solenoids} \right)}}{{length\;of\;wire}}\) 
• From given formula of self inductance, it is clear that it does not depends on the thickness of wire of solenoids, hence it is remain same.

ii) Rate of Joule heating if the same current goes through them:

Joule heating, H = i² Rt

And we know that, \(R = \rho \frac{l}{A} \Rightarrow R \propto \frac{1}{A}\) 

Since, A₁ > A₂, then R₁ < R₂ ⇒ H₁ < H₂

⇒ Joule heating of two solenoids is different for same current.

iii) Magnetic field energy if the same current goes through them:

Magnetic field energy - \(\left( M \right) = \frac{1}{2}L{i^2}\) 

⇒ M₁ = M₂

As the magnetic field does not depends on the thickness of the wire. Hence this energy is same for both wire.

iv) Time constant (τ):

\(\tau = \frac{L}{R} \Rightarrow \tau \propto \frac{1}{R}\;and\;R \propto \frac{1}{A}\)

A₁ > A₂ ⇒ R₁ < R₂ and hence τ₁ > τ₂

Therefore time constant of both the solenoid is different as it depends on the resistance of solenoid which is ultimately depends on the area of wire.