Question

A rectangular loop \((4\,\text{cm} \times 5\,\text{cm})\) of wire carrying current of 0.25 A is placed in a magnetic field \(\vec{B} = [(0.20\,\text{T})\hat{j} + (0.50\,\text{T})\hat{k}]\) such that outward unit vector normal to the plane of the loop is \((0.80\,\hat{i} - 0.60\,\hat{j})\).

Calculate:

(a) the magnetic moment \(\vec{m}\) of the loop, and

(b) the torque \(\vec{\tau}\) acting on the loop.

Hint:

Find the area in square metres, then m = I A n. Take the cross product m x B using the component rules for i, j and k.

Answer 1

Step 1: Understanding the concept:
A current loop of area \(A\) carrying current \(I\) has magnetic moment \(\vec{m} = IA\,\hat{n}\), where \(\hat{n}\) is the unit normal. In a field \(\vec{B}\) the torque is \(\vec{\tau} = \vec{m}\times\vec{B}\).

Step 2: (a) Magnetic moment:
\[ A = 4\ \text{cm}\times 5\ \text{cm} = 20\ \text{cm}^{2} = 2\times 10^{-3}\ \text{m}^{2} \]
\[ IA = 0.25\times 2\times 10^{-3} = 5\times 10^{-4}\ \text{A m}^{2} \]
\[ \vec{m} = 5\times 10^{-4}\,(0.80\hat{i} - 0.60\hat{j}) = (4\times 10^{-4}\hat{i} - 3\times 10^{-4}\hat{j})\ \text{A m}^{2} \]

Step 3: (b) Torque on the loop:
\[ \vec{\tau} = \vec{m}\times\vec{B} = (4\times 10^{-4}\hat{i} - 3\times 10^{-4}\hat{j})\times(0.20\hat{j} + 0.50\hat{k}) \]
Use \(\hat{i}\times\hat{j} = \hat{k}\), \(\hat{i}\times\hat{k} = -\hat{j}\), \(\hat{j}\times\hat{j} = 0\) and \(\hat{j}\times\hat{k} = \hat{i}\).
\[ \vec{\tau} = 8\times 10^{-5}\hat{k} - 2\times 10^{-4}\hat{j} + 0 - 1.5\times 10^{-4}\hat{i} \]
\[ \vec{\tau} = (-1.5\times 10^{-4}\hat{i} - 2.0\times 10^{-4}\hat{j} + 0.8\times 10^{-4}\hat{k})\ \text{N m} \]
Its size is \(\sqrt{(1.5)^{2} + (2.0)^{2} + (0.8)^{2}}\times 10^{-4} \approx 2.6\times 10^{-4}\ \text{N m}\).

Final Answer:
(a) \(\vec{m} = (4\hat{i} - 3\hat{j})\times 10^{-4}\) A m\(^2\). (b) \(\vec{\tau} = (-1.5\hat{i} - 2.0\hat{j} + 0.8\hat{k})\times 10^{-4}\) N m. \[ \boxed{\vec{\tau} = (-1.5\hat{i} - 2.0\hat{j} + 0.8\hat{k})\times 10^{-4}\ \text{N m}} \]