Question

A wire carrying current $I$ along x-axis has length $\ell$ and is kept in a magnetic field $\vec{B=(\hat{i}+2\hat{j}-3\hat{k})\,\text{Wb m}^{-2}$. The magnitude of magnetic force acting on the wire is}

• A. $\sqrt{15}\,I\ell B$
• B. $\sqrt{11}\,I\ell B$
• C. $\sqrt{13}\,I\ell B$
• D. $\sqrt{19}\,I\ell B$

Hint:

Magnetic force depends only on the component of $\vec{B}$ perpendicular to current.

Answer 1

The Correct Option is C

Step 1: Magnetic force on a current-carrying wire.
\[ \vec{F} = I(\vec{\ell} \times \vec{B}) \]
Step 2: Direction of length vector.
Since wire is along x-axis:
\[ \vec{\ell} = \ell \hat{i} \]
Step 3: Cross product.
\[ \vec{\ell} \times \vec{B} = \ell \hat{i} \times (\hat{i}+2\hat{j}-3\hat{k}) \] \[ = \ell(0 + 2\hat{k} + 3\hat{j}) \]
Step 4: Magnitude of force.
\[ |\vec{F}| = I\ell\sqrt{(3)^2 + (2)^2} = I\ell\sqrt{13} \]
Step 5: Conclusion.
The magnitude of magnetic force is $\sqrt{13}\,I\ell B$.