A) BHIl, 0
B) 2BIl, 0
C) 0, BIl
D) 0, 0
Correct Answer: C
Solution :
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 \[\theta =0,\] \[\therefore \] \[{{F}_{PQ}}=IBl\sin {{0}^{o}}=0\] Also, wire QR is perpendicular to the direction of magnetic field, then\[\theta ={{90}^{o}}\]. \[\therefore \] \[{{F}_{QR}}=IBl\sin {{90}^{o}}=IBl\]You need to login to perform this action.
You will be redirected in
3 sec