Question

A straight thin wire carrying a current of \(I=1\) A in the direction
\[ \hat{n} = \frac{1}{\sqrt6}\hat{i}+\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k} \]
is placed inside a magnetic field of \(\vec{B} = (2\hat{i}+2\hat{k})\) T.
The force per unit length on the wire is \(\_\_\_\_\_\) N/m.

• A. \(2\sqrt6\,\hat{i} - 2\sqrt6\,\hat{k}\)
• B. \(\sqrt{\dfrac{3}{2}}\,\hat{i} - \sqrt{\dfrac{3}{4}}\,\hat{j} + \sqrt{\dfrac{3}{4}}\,\hat{k}\)
• C. \(\dfrac{4}{\sqrt6}\,\hat{i} - \dfrac{4}{\sqrt6}\,\hat{k}\)
• D. \(\sqrt{\dfrac{3}{2}}\,\hat{i} + \sqrt{\dfrac{3}{4}}\,\hat{j} + \sqrt{\dfrac{3}{4}}\,\hat{k}\)

Hint:

Use \(\vec{f}=I(\hat{n}\times\vec{B})\) and expand the cross product component-wise; remember \(\hat{n}\) is already a unit vector.

Answer 1

The Correct Option is C

Instead of expanding the cross product as a 3 by 3 determinant, we can pull the normalizing factor out first and work with the plain direction vector, then divide by its length at the end. This uses the basic unit vector cross product rules: $\hat{i}\times\hat{i}=\hat{j}\times\hat{j}=\hat{k}\times\hat{k}=0$, $\hat{i}\times\hat{j}=\hat{k}$, $\hat{j}\times\hat{k}=\hat{i}$, $\hat{k}\times\hat{i}=\hat{j}$, and each reversed order flips the sign.

The plain direction vector before normalizing is $\vec{n}_0=\hat{i}+2\hat{j}+\hat{k}$, with length $|\vec{n}_0|=\sqrt{1^2+2^2+1^2}=\sqrt6$, so $\hat{n}=\vec{n}_0/\sqrt6$.

First find $\vec{n}_0\times\vec{B}$ where $\vec{B}=2\hat{i}+2\hat{k}$:

$$ \vec{n}_0\times\vec{B} = (\hat{i}+2\hat{j}+\hat{k})\times(2\hat{i}+2\hat{k}) $$

Expand term by term:

$$ = \hat{i}\times2\hat{i} + \hat{i}\times2\hat{k} + 2\hat{j}\times2\hat{i} + 2\hat{j}\times2\hat{k} + \hat{k}\times2\hat{i} + \hat{k}\times2\hat{k} $$

Using the rules above, $\hat{i}\times\hat{i}=0$, $\hat{i}\times\hat{k}=-\hat{j}$, $\hat{j}\times\hat{i}=-\hat{k}$, $\hat{j}\times\hat{k}=\hat{i}$, $\hat{k}\times\hat{i}=\hat{j}$, $\hat{k}\times\hat{k}=0$:

$$ = 0 - 2\hat{j} - 4\hat{k} + 4\hat{i} + 2\hat{j} + 0 = 4\hat{i} - 4\hat{k} $$

Now divide by $\sqrt6$ to bring back the unit vector, and multiply by the current $I=1$ A to get the force per unit length:

$$ \vec{f} = I(\hat{n}\times\vec{B}) = 1\times\frac{4\hat{i}-4\hat{k}}{\sqrt6} = \frac{4}{\sqrt6}\hat{i} - \frac{4}{\sqrt6}\hat{k}\ \text{N/m} $$

Let's summarize:

• Expanding the cross product term by term with the basic unit vector rules avoids setting up a determinant.
• Dividing the result by $\sqrt6$ at the end correctly accounts for the normalization of $\hat{n}$.
• The result matches option (C).

$$ \boxed{\vec{f} = \dfrac{4}{\sqrt6}\hat{i} - \dfrac{4}{\sqrt6}\hat{k}\ \text{N/m}} $$