Physics - Magnetism Question with Solution | TestHub

The magnetic force on a current-carrying wire is given by $\vec{F}=I(\vec{L}\times \vec{B})$.

Since the wire carries current along the $x$-axis, the length vector is $\vec{L}=L\hat{i}$.

The given magnetic field is $\vec{B}=B_0(\hat{i}-\hat{j}-\hat{k})$.

Substituting these into the force formula:

$\vec{F}=I(L\hat{i}\times B_0(\hat{i}-\hat{j}-\hat{k}))$

$\vec{F}=IL{B}_0(\hat{i}\times \hat{i}-\hat{i}\times \hat{j}-\hat{i}\times \hat{k})$

Using the properties of cross products of unit vectors ($\hat{i}\times \hat{i}=0$, $\hat{i}\times \hat{j}=\hat{k}$, $\hat{i}\times \hat{k}=-\hat{j}$):

$\vec{F}=IL{B}_0(0-\hat{k}-(-\hat{j}))=IL{B}_0(\hat{j}-\hat{k})$

The magnitude of the magnetic force is:

$|\vec{F}|=IL{B}_0\sqrt{(1{)}^2+(-1{)}^2}=\sqrt{2}IL{B}_0$

Answer: $\sqrt{2}{\mathrm{ILB}}_0$