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:
Torque ($\vec{\tau}$) on a current loop in a magnetic field is given by the cross product of its magnetic moment ($\vec{m}$) and the magnetic field ($\vec{B}$). The magnetic moment is the product of current, area, and the unit normal vector.
Step 2: Key Formula or Approach:
1. $\vec{m} = I \vec{A} = I (A \hat{n})$.
2. $\vec{\tau} = \vec{m} \times \vec{B}$.
Step 3: Detailed Explanation:
Area $A = \pi r^2 = \pi (0.02)^2 = 4\pi \times 10^{-4} \text{ m}^2$. Magnetic Moment: \[ \vec{m} = (100\sqrt{2})(4\pi \times 10^{-4}) \left(\frac{\hat{i} + \hat{k}}{\sqrt{2}}\right) = 4\pi \times 10^{-2} (\hat{i} + \hat{k}) \] Magnetic Field: \[ \vec{B} = 4 \times 10^{-3} (3\hat{i} + 2\hat{k}) = 10^{-3} (12\hat{i} + 8\hat{k}) \] Torque: \[ \vec{\tau} = (4\pi \times 10^{-2} (\hat{i} + \hat{k})) \times (10^{-3} (12\hat{i} + 8\hat{k})) \] \[ \vec{\tau} = 4\pi \times 10^{-5} [(\hat{i} \times 12\hat{i}) + (\hat{i} \times 8\hat{k}) + (\hat{k} \times 12\hat{i}) + (\hat{k} \times 8\hat{k})] \] \[ \vec{\tau} = 4\pi \times 10^{-5} [0 - 8\hat{j} + 12\hat{j} + 0] = 4\pi \times 10^{-5} (4\hat{j}) = 16\pi \times 10^{-5} \hat{j} \] \[ \vec{\tau} = 16 \times 3.14 \times 10^{-5} \hat{j} = 50.24 \times 10^{-5} \hat{j} = 5024 \times 10^{-7} \hat{j} \]
Step 4: Final Answer:
The torque experienced by the loop is \(5024 \times 10^{-7} \hat{j}\).