NEET Solved Paper 2016 Question 26
Question: A long straight wire of radius a carries a steady current $ I\text{.} $ The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B’, at radial distances $ \frac{a}{2} $ and 2a respectively, from the axis of the wire is :
Options:
A) $ \frac{1}{4} $
B) $ \frac{1}{2} $
C) 1
D) 4
Show Answer
Answer:
Correct Answer: C
Solution:
For points inside the wire $ B=\frac{{\mu_0}Ir}{2\pi R^{2}}(r\le R) $ For points outside the wire $ B=\frac{{\mu_0}I}{2\pi R}(r\ge R) $
according to the question $ \frac{B}{B’}=\frac{\frac{{\mu_0}I(a/2)}{2\pi a^{2}}}{\frac{{\mu_0}I}{2\pi (2a)}}=1:1 $
BETA
