A) 0.65
B) 0.55
C) 0.75
D) 0.45
Correct Answer: A
Solution :
Given, mass of the wire = 200 g = 02 kg Length of the wire = 1.5 m Current i = 2 A Magnitic field B = ? The force acting on the current carrying wire in uniform magnetic field \[F=Bil\sin \theta \] \[F=Bil\] \[(\because \,\theta ={{90}^{o}})\] Weight of the wire \[w=mg=0.2\times 9.8N\] In the position of suspension \[Bil=mg\] \[B=\frac{mg}{il}=\frac{0.2\times 9.8}{2\times 1.5}=0.65\,T\]
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