Magnetic Effects Of Current Question 80
Question: The magnetic field at the centre of coil of n turns, bent in the form of a square of side 2l , carrying current i, is [AMU (Engg.) 1999]
Options:
A) $ \frac{\sqrt{2}{\mu_{0}}nI}{\pi l} $
B) $ \frac{\sqrt{2}{\mu_{0}}nI}{2\pi l} $
C) $ \frac{\sqrt{2}{\mu_{0}}nI}{4\pi l} $
D) $ \frac{2{\mu_{0}}nI}{\pi l} $
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Answer:
Correct Answer: A
Solution:
Magnetic field due to one side of the square at centre O $ B_{1}=\frac{{\mu_{0}}}{4\pi }.\frac{2i\sin 45^{o}}{a/2} $
Þ $ B_{1}=\frac{{\mu_{0}}}{4\pi }.\frac{2\sqrt{2},i}{a} $ Hence magnetic field at centre due to all side $ B=4B_{1}=\frac{{\mu_{0}}(2\sqrt{2}i)}{\pi a} $ Magnetic field due to n turns $ B_{net}=nB=\frac{{\mu_{0}}2\sqrt{2}ni}{\pi a}=\frac{{\mu_{0}}2\sqrt{2}ni}{\pi (2l)} $ $ =\frac{\sqrt{2}{\mu_{0}}ni}{\pi l} $ $ (\because ,a=2l) $
BETA
