Magnetic Effects Of Current Question 248
Question: A loop of flexible conducting wire of length $ l $ lies in magnetic field B which is normal to the plane of loop. A current is passed through the loop. The tension I developed in the wire to open up is
Options:
A) $ \frac{\pi }{2}BIl $
B) $ \frac{BIl}{2} $
C) $ \frac{BIl}{2\pi } $
D) $ BIl $
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Answer:
Correct Answer: C
Solution:
[c] $ BI,(dl)=2T\sin (d\theta /2) $
$ \Rightarrow BI(rd\theta )=2T(d\theta \text{/}2) $ ( $ \theta $ is small, $ (\sin \theta =\theta ) $
$ \Rightarrow T=BIr=BIl/2\pi $
BETA
