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Q.

 A wire in the form of semi circle lies on a friction less table. The ends of semi circle are attached to springs S1 and S2 whose other ends a and b are fixed. If the current in the wire is I then tension in each spring is

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a

πRIB

b

2RIB

c

 IRB 

d

Zero

answer is C.

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Detailed Solution

Fm=Bil(for arc ac   l=2R)

 tension in each spring

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Is Ts then

2Ts=Fm⇒2Ts=2BIR⇒Ts=BIR

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 A wire in the form of semi circle lies on a friction less table. The ends of semi circle are attached to springs S1 and S2 whose other ends a and b are fixed. If the current in the wire is I then tension in each spring is