A) \[0.\text{65}\]
B) \[0.\text{55}\]
C) \[0.\text{75}\]
D) \[0.\text{45}\]
Correct Answer: A
Solution :
Given, mass of the wire = 200 g = 02 kg Length of the wire = 15 m Current \[i=2A\] Magnitic field B =? The force acting on the current carrying wire in uniform magnetic field \[F=Bil\sin \theta \] \[F=Bil\] \[(\because \theta ={{90}^{\circ }})\] Weight of the wire w = mg = 0.2 \[\times \] 9.8N In the position of suspension \[Bil=mg\] \[B=\frac{mg}{il}=\frac{0.2\times 9.8}{2\times 1.5}=0.65T\]You need to login to perform this action.
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