A) \[BIl,\,\,0\]
B) \[BIl,\,\,0\]
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 magnetic field, then\[\theta =0\], \[\therefore \] \[{{F}_{PQ}}=IBl\sin {{0}^{o}}=0\] Also, wire \[QR\] is perpendicular to the direction 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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