(i) The magnitude and direction of force on the element AC of the wire is \[\frac{\sqrt{3}}{2}\] ILB directed into the page |
(ii) The magnitude and direction of force on the element AB of the wire is \[\frac{\sqrt{3}}{2}\] ILB directed into the page |
(iii) The total force on the loop ABCA is zero |
A) (i) and (ii)
B) (ii) only
C) (i) and (iii)
D) (ii) and (iii)
Correct Answer: B
Solution :
Force \[F=Bil\sin \theta \] \[=Bil\sin \theta \] \[=\frac{\sqrt{3}}{2}Bil\] From Fleming left hand rule, the magnitude and direction of force on the element AB of the wire is \[\frac{\sqrt{3}}{2}\] Bil directed into the page.You need to login to perform this action.
You will be redirected in
3 sec