Q. Uniform magnetic $B$ is directed vertically upwards and 3 wires of equal length $L$, carrying equal current $i$ are lying in a horizontal plane such that the first one is along north, second one along north-east and the third one at $60{^\circ}$ north of east. Force exerted by magnetic field $B$ on them is
A) Zero on the first
B) $\frac{BiL}{\sqrt{2}}$ on the second
C) $\sqrt{3}\frac{BiL}{2}$ on the third
D) $BiL$ on all of them
Explanation
(d) $F = I.(L_{eff})B$
$L_{eff} = L + L\cos{45{^\circ}} + L\sin{60{^\circ}}$
$= L + L \times \frac{1}{\sqrt{2}} + L \times \frac{\sqrt{3}}{2}$
$= L\left( \frac{2 + \sqrt{2} + \sqrt{3}}{2} \right)$
Force on each wire $= I.L.B\sin\theta$ [Because angle between current element and magnetic field $= 90{^\circ}$]
$= I.L.B.\sin{90{^\circ}} = I.L.B.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.