Question

A loop ABCD, carrying current $ I = 12 \, \text{A} $, is placed in a plane, consists of two semi-circular segments of radius $ R_1 = 6\pi \, \text{m} $ and $ R_2 = 4\pi \, \text{m} $. The magnitude of the resultant magnetic field at center O is $ k \times 10^{-7} \, \text{T} $. The value of $ k $ is ______ (Given $ \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} $)

Hint:

Use the formula for the magnetic field due to a circular arc at its center. Remember to consider the direction of the magnetic field due to each segment and subtract them.

Answer 1

Correct Answer: 1

This problem requires determining the resultant magnetic field at point O, the center of a loop ABCDA. The loop comprises two semi-circular segments with radii R₁ and R₂ and two straight segments. Given the current I, radii R₁, R₂, and the constant μ₀, the objective is to find the value of k, where the magnetic field is expressed as k × 10⁻⁷ T.

Concept Used:

The magnetic field at the center of a current-carrying circular arc is calculated using the formula:

\[ B = \frac{\mu_0 I \theta}{4\pi R} \]

Here, I represents the current, R is the radius, θ is the angle subtended by the arc at the center in radians, and μ₀ is the permeability of free space. For a semi-circular arc, θ = π radians, simplifying the formula to:

\[ B_{semi-circle} = \frac{\mu_0 I}{4R} \]

The magnetic field produced by a straight current-carrying wire at any point along its length is zero. The direction of the magnetic field is determined by the right-hand thumb rule. The net magnetic field at a point due to multiple sources is the vector sum of the individual fields, as per the Principle of Superposition.

Step-by-Step Solution:

Step 1: Evaluate the magnetic field generated by each segment of the loop ABCDA at the center O.

The loop is composed of four segments: AB, BC, CD, and DA.

• Segments AB and CD: These are straight wires situated on the line passing through center O. Consequently, the magnetic field at O due to these segments is zero.
• Segment BC: This is a semi-circular arc with radius R₁. Assuming this is the inner arc, the current flows from B to C (clockwise). Applying the right-hand thumb rule, the magnetic field at O from this segment (B₁) is directed into the plane of the paper.
• Segment DA: This is a semi-circular arc with radius R₂. Assuming this is the outer arc, the current flows from D to A (counter-clockwise). Applying the right-hand thumb rule, the magnetic field at O from this segment (B₂) is directed out of the plane of the paper.

Step 2: Compute the magnitudes of the magnetic fields from the semi-circular segments.

The problem specifies R₁ = 6π m and R₂ = 4π m. The diagram indicates R₁ as the inner radius and R₂ as the outer radius. However, the given values (R₁ = 6π m, R₂ = 4π m) imply R₂<R₁. This contradicts the visual representation of R₂ being the outer radius. Adhering to the textual values for R₁ and R₂ and the current path ABCDA, the arc BC has radius R₁ = 6π m, and the magnetic field at O is:

\[ B_1 = \frac{\mu_0 I}{4R_1} \quad \text{(into the page)} \]

The arc DA has radius R₂ = 4π m, and the magnetic field at O is:

\[ B_2 = \frac{\mu_0 I}{4R_2} \quad \text{(out of the page)} \]

Step 3: Calculate the resultant magnetic field at the center O.

The fields B₁ and B₂ are in opposing directions. The net magnetic field B_net is the difference between their magnitudes. Since R₂<R₁, it follows that B₂>B₁.

\[ B_{net} = B_2 - B_1 \]

\[ B_{net} = \frac{\mu_0 I}{4R_2} - \frac{\mu_0 I}{4R_1} = \frac{\mu_0 I}{4} \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \]

The net field is directed out of the page.

Step 4: Substitute the given values into the expression for the net magnetic field.

Given:

• \( I = 12 \, \text{A} \)
• \( R_1 = 6\pi \, \text{m} \)
• \( R_2 = 4\pi \, \text{m} \)
• \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \)

\[ B_{net} = \frac{(4\pi \times 10^{-7}) \times 12}{4} \left( \frac{1}{4\pi} - \frac{1}{6\pi} \right) \]

\[ B_{net} = (12\pi \times 10^{-7}) \left( \frac{3 - 2}{12\pi} \right) \]

\[ B_{net} = (12\pi \times 10^{-7}) \left( \frac{1}{12\pi} \right) \]

\[ B_{net} = 1 \times 10^{-7} \, \text{T} \]

 

Step 5: Determine the value of k.

The problem states that the magnitude of the resultant magnetic field at center O is \( k \times 10^{-7} \, \text{T} \). Comparing this with our calculated result:

\[ 1 \times 10^{-7} = k \times 10^{-7} \]

\[ k = 1 \]

The value of k is 1.