Torque on a Current Loop — Question 3

Question 3

A three-dimensional area normal fixes the full torque vector

A single-turn loop has area .0200 square metre and current 5.00 A. Its current-defined normal is $\hat n=(2\hat x-\hat y+2\hat z)/3$. The applied field is $.300\hat y$ T.

1. Find the torque vector and magnitude.

2. Find the angle between the normal and the field and verify the torque magnitude using the sine formula.

Question 3 - Solution

A three-dimensional area normal fixes the full torque vector

(a) Method and calculation

$$\vec\tau=IA\hat n\times\vec B=\frac{(5)(.0200)(.300)}3(-2\hat x+2\hat z).$$$$\vec\tau=(-.0200\hat x+.0200\hat z)\,\mathrm{N\,m},\quad\tau=.0200\sqrt2\,\mathrm{N\,m}.$$

(b) Interpretation and further calculation

$$\cos\theta=\hat n\cdot\hat y=-1/3\Rightarrow\theta=\cos^{-1}(-1/3).$$$$IA B\sin\theta=.0300\sqrt{1-1/9}=.0200\sqrt2\,\mathrm{N\,m}.$$

Physical check

The negative normal component along B makes the angle obtuse, but the torque magnitude remains positive.