Q. A wire $PQR$ is bent as shown in figure and is placed in a region of uniform magnetic field $B.$ The length of $PQ = QR = l.$ A current $I$ ampere flows through the wire as shown. The magnitude of the force on $PQ\ \ and\ \ QR$ will be
A) $BIl,\ 0$
B) $2BIl,\ 0$
C) $0,\ BIl$
D) $0,\ 0$
Explanation
(c) The Lorentz force acting on the current carrying conductor in the magnetic field is
$F = IBl\sin\theta$
Since, wire $PQ$ is parallel to the direction of magnetic field, then $\theta = 0,$
$\therefore F_{PQ} = IBl\sin{0{^\circ}} = 0$
Also, wire $QR$ is perpendicular to the direction of magnetic field, then $\theta = 90{^\circ}.$
$\therefore F_{QR} = IBl\sin{90{^\circ}} = IBl$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.