Magnetic Effects Of Current Question 428
Question: The magnetic field at O due to current in the infinite wire forming a loop as shown in Fig. is
Options:
A) $ \frac{{\mu_{0}}I}{2\pi d}(\cos {\phi_{1}}+\cos {\phi_{2}}) $
B) $ \frac{{\mu_{0}}I2I}{4\pi d}(tan{\theta_{1}}+\tan {\theta_{2}}) $
C) $ \frac{{\mu_{0}}I}{4\pi d}(sin{\phi_{1}}+\sin {\phi_{2}}) $
D) $ \frac{{\mu_{0}}I}{4\pi d}(cos{\theta_{1}}+\sin {\theta_{2}}) $
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Answer:
Correct Answer: A
Solution:
[a] Using $ B=\frac{{\mu_{0}}I}{4\pi d}[ \sin ,{\theta_{1}}+\sin ,{\theta_{2}} ] $ But $ {\theta_{1}}+{\phi_{1}}=90^{o} $ or $ {\theta_{1}}=90^{o}-\phi $ , $ \sin {\theta_{1}}=\sin (90^{o}-{\phi_{1}})=\cos {\phi_{1}} $ Similarly, $ \sin ,{\theta_{2}}=\cos ,{\phi_{2}} $ $ B_{net}=\frac{{\mu_{0}}I}{2\pi d}( \cos ,{\phi_{1}}+\cos ,{\phi_{2}} ) $
BETA
