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

Step 1: Recall the force law for a current-carrying wire in a magnetic field.
For a straight wire carrying current \(I\) along direction \(\hat{n}\), placed in a field \(\vec{B}\), the force per unit length is
\[ \vec{f} = I(\hat{n}\times\vec{B}) \]

Step 2: Write down the given vectors.
\[ \hat{n} = \frac{1}{\sqrt6}\hat{i}+\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}, \qquad \vec{B}=2\hat{i}+0\hat{j}+2\hat{k}, \qquad I=1\ \text{A} \]

Step 3: Compute the cross product \(\hat{n}\times\vec{B}\) using the determinant form.
\[ \hat{n}\times\vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{1}{\sqrt6} & \frac{2}{\sqrt6} & \frac{1}{\sqrt6} \\ 2 & 0 & 2 \end{vmatrix} \]
The \(\hat{i}\) component: \(\left(\frac{2}{\sqrt6}\times2\right)-\left(\frac{1}{\sqrt6}\times0\right) = \frac{4}{\sqrt6}\).
The \(\hat{j}\) component: \(-\left[\left(\frac{1}{\sqrt6}\times2\right)-\left(\frac{1}{\sqrt6}\times2\right)\right] = 0\).
The \(\hat{k}\) component: \(\left(\frac{1}{\sqrt6}\times0\right)-\left(\frac{2}{\sqrt6}\times2\right) = -\frac{4}{\sqrt6}\).

Step 4: Write the force per unit length.
Since \(I=1\) A, the force per unit length equals the cross product itself:
\[ \vec{f} = \frac{4}{\sqrt6}\hat{i} - \frac{4}{\sqrt6}\hat{k}\ \text{N/m} \]

Step 5: Check the other options.
Option (A) would come from using the un-normalized direction \((1,2,1)\), of magnitude \(\sqrt6\), in place of the unit vector \(\hat{n}\), which overstates the force by a factor of 6. Options (B) and (D) do not come from a correctly worked cross product, they mix up which components survive and which cancel, and option (D) even keeps a positive \(\hat{k}\) term where the correct working gives a negative one.

Final Answer:
\[ \boxed{\vec{f} = \frac{4}{\sqrt6}\hat{i} - \frac{4}{\sqrt6}\hat{k}\ \text{N/m}} \]