Question

A straight wire carrying current $5A$ lies along the body diagonal of an imaginary cube of side $20cm$. If a uniform magnetic field of $0.6T$ exists perpendicular to one of the faces of cube, then the force on the wire in Newtons is (Take $\sqrt{2}=1.41$ and write up to two digits after the decimal point.)

Answer 1

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}\Rightarrow F=ilB\sin \alpha$
here $l=\sqrt{3}a$ & from the diagram, $\cos \alpha =\frac{a}{\sqrt{3a}}=\frac{1}{\sqrt{3}}$
So, $\sin \alpha =\frac{\sqrt{2}}{\sqrt{3}}$
Therefore, $F=i\sqrt{3}aB\sqrt{\frac{2}{3}}=\sqrt{2}iaB$
$=1.41\times 5\times 0.2\times 0.6=0.846N$