A) \[BIl,0\]
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.
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