Magnetic Effects Of Current Question 419
Question: If the magnetic field at P can be written as K tan $ ( \frac{\alpha }{2} ) $ , then K
Options:
A) $ \frac{{\mu_{0}}I}{4\pi d} $
B) $ \frac{{\mu_{0}}I}{2\pi d} $
C) $ \frac{{\mu_{0}}I}{\pi d} $
D) $ \frac{2{\mu_{0}}I}{\pi d} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let us compute the magnetic field due to any one segment: $ B=\frac{{\mu_{0}}i}{4\pi (d,\sin ,\alpha )}( \cos ,0^{o}+\cos ,(180-\alpha ) ) $ $ =\frac{{\mu_{0}}I}{4\pi (d,\sin ,\alpha )}(1-\cos \alpha ) $ $ =\frac{{\mu_{0}}I}{4\pi d}\tan ,\frac{\alpha }{2} $ Resultant field will be $ B_{net}=2B=\frac{{\mu_{0}}I}{2\pi d}\tan \frac{\alpha }{2}\Rightarrow K=\frac{{\mu_{0}}I}{2\pi d} $
BETA
