Magnetic Field due to a Current Element MCQ Quiz in मल्याळम - Objective Question with Answer for Magnetic Field due to a Current Element - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT:

• Magnetic field: The space or region around the current-carrying wire/moving electric charge or around the magnetic material in which force of magnetism can be experienced by other magnetic material is called as magnetic field/magnetic induction by that material/current.
  • It is denoted by B and the SI unit of the magnetic field is the tesla (T).
• The magnetic field at the center of a circular coil is given by:

$\Rightarrow B=\frac{{\mu }_{0}I}{2R}$

Where B = magnetic field at the center of the coil, μ0 is the permeability of free space = 4π × 10-7, I is current and R is the radius of the circular coil.

• When there are n turns in the coil then the magnetic field produced by the coil at the center is given by:

$\Rightarrow B^{\prime }=\frac{{\mu }_{0}nI}{2R}$

​CALCULATION:

Given I = 2 amp, N = 1 and B = 4π×10^-6 Wb/m²

• The magnetic field produced at the center of the current-carrying coil is given as,

$\Rightarrow B=\frac{{\mu }_{o}nI}{2r}$     -----(1)

Where r = radius of the coil and μ_o = 4π×10^-7 H/m

By equation 1,

$\Rightarrow r=\frac{{\mu }_{o}nI}{2B}$

$\Rightarrow r=\frac{4\pi \times {10}^{-7}\times 1\times 2}{2\times 4\pi \times {10}^{-6}}$

$\Rightarrow r=0.1m$

Calculation:

(A) 

B_ab = (μ₀ / 4π) × (I / r) (out of the plane)

B_bcd = (μ₀ / 4π) × (I / r) × (2π) (in the plane)

B_de = (μ₀ / 4π) × (I / r) (out of the plane)

Hence magnetic field at O is

B₀ = –(μ₀ / 4π) × (I / r) + (μ₀ / 4π) × (I / r) × 2π – (μ₀ / 4π) × (I / r)

B₀ = (μ₀ / 2π) × (I / r) × (π – 1) ...(III)

(B) 

B_ab = (μ₀ / 4π) × (I / r) (out of the plane)

B_bcd = (μ₀ / 4π) × (I / r) × π (out of the plane)

B_de = (μ₀ / 4π) × (I / r) (out of the plane)

Hence magnetic field at O is

B₀ = (μ₀ / 4π) × (I / r) + (μ₀ / 4π) × (I / r) × π + (μ₀ / 4π) × (I / r)

B₀ = (μ₀ / 4π) × (I / r) × (π + 2) ...(I)

(C) 

B_ab = (μ₀ / 4π) × (I / r) (in the plane)

B_bcd = (μ₀ / 4π) × (I / r) × π (in the plane)

B_de = 0 (at the axis)

Hence magnetic field at O is

B₀ = (μ₀ / 4π) × (I / r) × (1 + π) ...(IV)

(D) 

B_ab = 0 (at the axis)

B_bcd = (μ₀ / 4π) × (I / r) × π (out of the plane)

B_de = 0 (at the axis)

Hence magnetic field at O is

B₀ = (μ₀ × I) / (4r) ...(II)

Magnetic field at centre at coil $B_1={\displaystyle \frac{{\mu }_{0}i}{2a}}$

Magnetic field at distance a from centre $B_2={\displaystyle \frac{{\mu }_{0}i}{2a}}{\sin }^3\theta$

Refer image,

$\sin \theta ={\displaystyle \frac{1}{2\sqrt{2}}}/{\sin }^3\theta ={\displaystyle \frac{1}{2\sqrt{2}}}$

$B_2={\displaystyle \frac{{\mu }_{0}i}{2a}}\left({\displaystyle \frac{1}{2\sqrt{2}}}\right)$

${\displaystyle \frac{B_1}{B_2}}={\displaystyle \frac{{\displaystyle \frac{{\mu }_{0}i}{2a}}}{{\displaystyle \frac{{\mu }_{0}i}{2a}}\left({\displaystyle \frac{1}{2\sqrt{2}}}\right)}}=2\sqrt{2}$

$\boxed{{\displaystyle {\displaystyle \frac{B_1}{B_2}}=2\sqrt{2}}}$

The correct answer is Option 3.

Concept:

Magnetic field due to straight conductor: (μ₀I/4πR) × (sinθ₁ + sinθ₂)

Magnetic field due to circular conductor: (μ₀I/4πR) × θ (in radians)

Direction determined by right-hand thumb rule.

Calculation:

Given:

Semi-circular conductor (angle = 270° = 3π/2 radians)

Straight sections forming 90° at point P

Magnetic field due to semi-circular wire: (μ₀I/4πR) × (3π/2)

Magnetic field due to each straight conductor: (μ₀I/4πR) × sin(90°) = (μ₀I/4πR) × (1) (Assuming the wire to be long)

For two straight wires: 2 × (μ₀I/4πR) × (1) = (μ₀I/4πR) × 2

Net magnetic field: Semi-circular part field – Straight part field = (μ₀I/4πR) × (3π/2 – 2)

Explanation:

Magnetic Field Near a Long, Straight Current-Carrying Conductor

Definition: The magnetic field at a point near a long, straight current-carrying conductor is the region around the conductor where magnetic forces can be observed. The strength and direction of this magnetic field depend on the current flowing through the conductor and the distance from the conductor.

Working Principle: According to Ampère's Law and the Biot-Savart Law, the magnetic field (B) around a long, straight conductor carrying current (I) is directly proportional to the current and inversely proportional to the distance (r) from the conductor. Mathematically, this relationship is expressed as:

B = (μ₀ × I) / (2π × r)

where:

• B = Magnetic field strength
• μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
• I = Current through the conductor
• r = Distance from the conductor

Correct Option Analysis:

The correct option is:

Option 4: The current through the conductor

This option is correct because the magnetic field (B) is directly proportional to the current (I) through the conductor. As the current increases, the magnetic field strength also increases, and vice versa. The proportionality is linear, meaning that if the current doubles, the magnetic field strength also doubles

Concept:

The magnetic field on the axis of a circular coil of radius R, carrying current I, is given by:

$B=\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi I{R}^2}{(R^2+x^2{)}^{3/2}}$

At the center of the coil (x = 0), the magnetic field is:

$B_{\text{center}}=\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi I}{R}$

At a distance x on the axis, the magnetic field is reduced to a fraction of the field at the center.

Calculation:

Given:

• $B_{\text{center}}=64\cdot B_x$
• $B_x=\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi I{R}^2}{(R^2+x^2{)}^{3/2}}$

Equating the ratio of fields:

$$\begin{gathered} \Rightarrow \frac{\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi I}{R}}{\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi I{R}^2}{(R^2+x^2{)}^{3/2}}}=64 \\ \Rightarrow \frac{(R^2+x^2{)}^{3/2}}{R^3}=64 \end{gathered}$$

Solving for x:

$$\begin{gathered} \Rightarrow (R^2+x^2{)}^{3/2}=64R^3 \\ \Rightarrow R^2+x^2=(64R^3{)}^{2/3}=16R^2 \\ \Rightarrow x^2=16R^2-R^2=15R^2 \\ \Rightarrow x=\sqrt{15}R \end{gathered}$$

∴ The distance $x=\sqrt{15}R.$

The correct options is 4).

Calculation: 

⇒ $\frac{dF}{dl}=\frac{{\mu }_0{i}_1{i}_2}{2\pi d}$

So

⇒ $\frac{2\times {10}^{-7}\times 5\times 5}{d}=\frac{{10}^{-5}}{10\times {10}^{-2}}$

⇒ $d=\frac{2\times {10}^{-7}\times 5\times 5}{{10}^{-4}}$

⇒ d = 50 mm = 5 cm

∴ the separation between the wires is 5 cm.

Calculation:

Magnetic field due to current carrying coil on axis at distance d.

⇒ B_a = $\frac{{\mu }_0\mathrm{I}{r}^2}{2{(r^2+d^2)}^{\frac{3}{2}}}$

Given that d = r

Now,

⇒ B_a = $\frac{{\mu }_0\mathrm{I}{r}^2}{2{\left(2r^2\right)}^{\frac{3}{2}}}=\frac{{\mu }_0\mathrm{I}}{4\sqrt{2}r}$

Magnetic field at centre of current carrying coil

⇒ ${\mathrm{B}}_c=\frac{{\mu }_0\mathrm{I}}{2r}$

⇒ $\frac{{\mathrm{B}}_c}{{\mathrm{B}}_a}=\frac{{\mu }_0\mathrm{I}}{2r}\times \frac{4\sqrt{2}r}{{\mu }_0\mathrm{I}}=\frac{2\sqrt{2}}{1}=\frac{\sqrt{8}}{1}$

x = 8

∴ the value of x is 8.

CONCEPT: 

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

B = $\frac{{\mu }_{0}I}{2\pi d}$

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 = $\frac{{\mu }_{0}I}{2\pi r}$

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

The correct answer is option 2.

CONCEPT:

Magnetic field due to a straight current-carrying wire:

• Magnetic field due to a straight current-carrying wire at a distance r is given as:

$\Rightarrow B=\frac{{\mu }_{o}I}{4\pi r}(sin{\theta }_1+sin{\theta }_2)$

CALCULATION:

Given I = a, r = a/2, and θ₁ = θ₂ = 45°

• So the magnetic field at the center of the given square loop is:

$\Rightarrow B=4\times \frac{{\mu }_{o}I}{4\pi \times \frac{a}{2}}(sin45+sin45)$

$\Rightarrow B=\frac{2\sqrt{2}{\mu }_{o}I}{\pi a}$

• Hence, option 1 is correct.