Question

A rectangular loop \((1\,\text{cm} \times 5\,\text{cm})\) of wire carrying current of 0.4 A is placed in a magnetic field \(\vec{B} = (0.03\hat{i} + 0.04\hat{k})\) T such that the outward unit vector normal to the plane of the loop is \((0.60\hat{i} - 0.80\hat{j})\). Calculate: (a) the magnetic moment \(\vec{m}\) of the loop, and (b) the torque \(\vec{\tau}\) acting on the loop.

Hint:

Use \(\vec{m} = IA\hat{n}\) with \(A = 5 \times 10^{-4}\ \text{m}^2\), then \(\vec{\tau} = \vec{m}\times\vec{B}\). The torque comes out as \((-6.4\hat{i} - 4.8\hat{j} + 4.8\hat{k}) \times 10^{-6}\ \text{N m}\).

Answer 1

Step 1: Understanding the Concept:
A current loop behaves like a magnetic dipole. Its magnetic moment points along the outward normal of the loop, and the torque on it in a field is the cross product of this moment with the field.

Step 2: Key Formula or Approach:
1. \(\vec{m} = I A \hat{n}\), where \(A\) is the area of the loop and \(\hat{n}\) is the outward unit normal.
2. \(\vec{\tau} = \vec{m} \times \vec{B}\).

Step 3: Find the magnetic moment:
Area of the loop: \(A = 1\ \text{cm} \times 5\ \text{cm} = 5\ \text{cm}^2 = 5 \times 10^{-4}\ \text{m}^2\).
Size of the moment: \(IA = 0.4 \times 5 \times 10^{-4} = 2 \times 10^{-4}\ \text{A m}^2\).
Multiply by the unit normal \(\hat{n} = 0.60\hat{i} - 0.80\hat{j}\):
\[ \vec{m} = 2 \times 10^{-4}\,(0.60\hat{i} - 0.80\hat{j}) = (1.2 \times 10^{-4}\hat{i} - 1.6 \times 10^{-4}\hat{j})\ \text{A m}^2 \]
Its magnitude is \(2 \times 10^{-4}\ \text{A m}^2\), because \(\hat{n}\) is a unit vector.

Step 4: Find the torque:
Take the cross product with \(\vec{B} = (0.03\hat{i} + 0.04\hat{k})\ \text{T}\):
\[ \vec{\tau} = 2 \times 10^{-4}\,\left[(0.60\hat{i} - 0.80\hat{j}) \times (0.03\hat{i} + 0.04\hat{k})\right] \]
Expand the four products using \(\hat{i}\times\hat{i} = 0\), \(\hat{i}\times\hat{k} = -\hat{j}\), \(\hat{j}\times\hat{i} = -\hat{k}\), \(\hat{j}\times\hat{k} = \hat{i}\):
\[ 0.60 \times 0.03\,(\hat{i}\times\hat{i}) = 0 \]
\[ 0.60 \times 0.04\,(\hat{i}\times\hat{k}) = -0.024\hat{j} \]
\[ -0.80 \times 0.03\,(\hat{j}\times\hat{i}) = +0.024\hat{k} \]
\[ -0.80 \times 0.04\,(\hat{j}\times\hat{k}) = -0.032\hat{i} \]
Add them: \((-0.032\hat{i} - 0.024\hat{j} + 0.024\hat{k})\). Multiply by \(2 \times 10^{-4}\):
\[ \vec{\tau} = (-6.4 \times 10^{-6}\hat{i} - 4.8 \times 10^{-6}\hat{j} + 4.8 \times 10^{-6}\hat{k})\ \text{N m} \]
Its magnitude is \(2 \times 10^{-4} \times \sqrt{0.032^2 + 0.024^2 + 0.024^2} \approx 9.3 \times 10^{-6}\ \text{N m}\).

Final Answer:
The magnetic moment is \(\vec{m} = (1.2 \times 10^{-4}\hat{i} - 1.6 \times 10^{-4}\hat{j})\ \text{A m}^2\) and the torque is \(\vec{\tau} = (-6.4\hat{i} - 4.8\hat{j} + 4.8\hat{k}) \times 10^{-6}\ \text{N m}\). \[ \boxed{\vec{m} = (1.2\hat{i} - 1.6\hat{j}) \times 10^{-4}\ \text{A m}^2,\quad \vec{\tau} = (-6.4\hat{i} - 4.8\hat{j} + 4.8\hat{k}) \times 10^{-6}\ \text{N m}} \]