A) \[2\sqrt{2}N{{m}^{-1}}\]
B) \[\frac{2}{\sqrt{2}}N{{m}^{-1}}\]
C) \[\frac{\sqrt{2}}{2}N{{m}^{-1}}\]
D) \[4\sqrt{2}N{{m}^{-1}}\]
Correct Answer: B
Solution :
(b) \[\frac{2}{\sqrt{2}}N{{m}^{-1}}\] \[I=10\,A,\,\theta =45{}^\circ ,B=0.2\,T\] \[\therefore F=lIB\,\sin \theta \] \[\therefore \frac{F}{l}=IB\,\sin \,45{}^\circ =10\times 0.2\times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}N{{m}^{-1}}\]You need to login to perform this action.
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