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 calculating the resultant magnetic field at the center O of a loop ABCDA. The loop consists of two semi-circular segments with radii R₁ and R₂ and two straight segments. We are given the current I, the radii R₁ and R₂, and the value of μ₀, and we need to find the value of k where the magnetic field is k × 10⁻⁷ T.

Concept Used:

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

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

where \( I \) is the current, \( R \) is the radius, \( \theta \) is the angle subtended by the arc at the center (in radians), and \( \mu_0 \) is the permeability of free space. For a semi-circular arc, \( \theta = \pi \) radians, so the formula simplifies to:

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

The magnetic field due to 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 (Principle of Superposition).

Step-by-Step Solution:

Step 1: Analyze the magnetic field due to each segment of the loop ABCDA at the center O.

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

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

 

Step 2: Calculate the magnitudes of the magnetic fields due to the semi-circular segments.

The problem text states R₁ = 6π m and R₂ = 4π m. The diagram labels the inner radius as R₁ and outer as R₂. Let's follow the labels in the diagram in conjunction with the direction of the current loop ABCDA. The arc BC has radius R₁. The current is clockwise. The arc DA has radius R₂. The current is counter-clockwise. From the given values, R₂ < R₁. This contradicts the visual representation in the diagram. Let's follow the textual values assigned to R₁ and R₂ and the current path ABCDA. The arc BC has radius R₁ = 6π m. 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. The magnetic field at O is: \[ B_2 = \frac{\mu_0 I}{4R_2} \quad \text{(out of the page)} \]

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

The two fields B₁ and B₂ are in opposite directions. The net magnetic field B_net is the difference between their magnitudes. Since \( R_2 < R_1 \), it follows that \( B_2 > B_1 \). \[ B_{net} = B_2 - B_1 \] \[ B_{net} = \frac{\mu_0 I}{4R_2} - \frac{\mu_0 I}{4R_1} = \frac{\mu_0 I}{4} \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \] The direction of the net field will be 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 our calculated result with the given expression: \[ 1 \times 10^{-7} = k \times 10^{-7} \] \[ k = 1 \]

The value of k is 1.

Answer 2

$B_0 = |B_{R_1} - B_{R_2}|$

$B_0 = \left| \frac{\mu_0 I}{4R_2} - \frac{\mu_0 I}{4R_1} \right|$

$B_0 = \frac{4\pi \times 10^{-7} \times 12}{4} \left| \frac{1}{4\pi} - \frac{1}{6\pi} \right|$

$B_0 = 12\pi \times 10^{-7} \left| \frac{1}{12\pi} \right|$

$B_0 = 1 \times 10^{-7} T$

$k = 1$