### 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 $