Magnetic Effects Of Current Question 436
Question: A current I flows in the anticlockwise direction through a square loop of side a lying in the XOY plane with its center at the origin. The magnetic induction at the center of the square loop is
Options:
A) $ \frac{2\sqrt{2}{\mu_{0}}I}{\pi a}{{\hat{e}}_{x}} $
B) $ \frac{2\sqrt{2}{\mu_{0}}I}{\pi a}{{\hat{e}}_{z}} $
C) $ \frac{2\sqrt{2}{\mu_{0}}I}{\pi a^{2}}{{\hat{e}}_{z}} $
D) $ \frac{2\sqrt{2}{\mu_{0}}I}{\pi a^{2}}{{\hat{e}}_{x}} $
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Answer:
Correct Answer: B
Solution:
[b] Field due to one side of loop at $ O=\frac{{\mu_{0}}I}{4\pi ( \frac{a}{2} )}(2,\sin ,45^{o}) $ Field at O due to all four sides is along unit vector $ {{\hat{e}}_{z}} $
$ \therefore $ Tool field $ =4.\frac{{\mu_{0}}I}{4\pi ( \frac{a}{2} )}(2,\sin ,45^{o})=\frac{2\sqrt{2}{\mu_{0}}I}{\pi a} $
BETA
