Question

A current carrying circular loop of radius 2 cm with unit normal \(\hat{n} = \frac{\hat{k} + \hat{i}}{\sqrt{2}}\) is placed in a magnetic field, \(\vec{B} = B_0 (3\hat{i} + 2\hat{k})\). If \(B_0 = 4 \times 10^{-3}\) T and current \(I = 100\sqrt{2}\) A, the torque experienced by the loop is ________ Wb·A.

• A. \(16 \times 10^{-5} \hat{k}\)
• B. \(5024 \times 10^{-7} \hat{k}\)
• C. \(5024 \times 10^{-7} \hat{i}\)
• D. \(5024 \times 10^{-7} \hat{j}\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A current-carrying loop in a magnetic field experiences a torque. We must first define the magnetic dipole moment vector of the loop using its area and normal vector, and then apply the cross product with the magnetic field vector.
Step 2: Key Formula or Approach:
1. Magnetic dipole moment: $\vec{m} = I \cdot A \cdot \hat{n}$.
2. Area of circular loop: $A = \pi r^2$.
3. Torque on a magnetic dipole: $\vec{\tau} = \vec{m} \times \vec{B}$.
Step 3: Detailed Explanation:
Given values:
Radius $r = 2\text{ cm} = 0.02\text{ m}$.
Area $A = \pi r^2 = 3.14 \times (0.02)^2 = 3.14 \times 0.0004 = 1.256 \times 10^{-3}\text{ m}^2$.
Current $I = 100\sqrt{2}\text{ A}$.
Unit normal vector $\hat{n} = \frac{\hat{i} + \hat{k}}{\sqrt{2}}$.
Calculate the magnetic dipole moment $\vec{m}$:
$\vec{m} = I \times A \times \hat{n} = (100\sqrt{2}) \times (1.256 \times 10^{-3}) \times \left( \frac{\hat{i} + \hat{k}}{\sqrt{2}} \right)$.
$\vec{m} = 100 \times 1.256 \times 10^{-3} (\hat{i} + \hat{k}) = 0.1256 (\hat{i} + \hat{k})\text{ A}\cdot\text{m}^2$.
The magnetic field is given by:
$\vec{B} = 4 \times 10^{-3} (3\hat{i} + 2\hat{k})\text{ T}$.
Now, calculate the torque $\vec{\tau} = \vec{m} \times \vec{B}$:
$\vec{\tau} = [0.1256 (\hat{i} + \hat{k})] \times [4 \times 10^{-3} (3\hat{i} + 2\hat{k})]$.
$\vec{\tau} = 0.1256 \times 4 \times 10^{-3} [(\hat{i} + \hat{k}) \times (3\hat{i} + 2\hat{k})]$.
Calculate the cross product terms:
$(\hat{i} \times 3\hat{i}) = 0$
$(\hat{i} \times 2\hat{k}) = -2\hat{j}$
$(\hat{k} \times 3\hat{i}) = 3\hat{j}$
$(\hat{k} \times 2\hat{k}) = 0$
Sum of cross product = $-2\hat{j} + 3\hat{j} = \hat{j}$.
Therefore:
$\vec{\tau} = (0.5024 \times 10^{-3}) \hat{j}$.
To match the options format, shift the decimal point:
$\vec{\tau} = 5024 \times 10^{-7} \hat{j}\text{ Wb.A}$.
Step 4: Final Answer:
The torque experienced is $5024 \times 10^{-7}\hat{j}$.