JEE Main On 08 April 2018 Question 24
Question: The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is $ B _1 $ . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is $ B _2. $ The ratio $ \frac{B _1}{B _2} $ is: [JEE Main Online 08-04-2018]
Options:
A) $ \sqrt{2} $
B) $ \frac{1}{\sqrt{2}} $
C) 2
D) $ \sqrt{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I(\pi r^{2})=M,2M=I\pi {{(r)}^{2}} $
$ r=\sqrt{2}r $ $ B _1=\frac{{\mu_o}I}{2r} $
$ B _2=\frac{{\mu_o}I}{2\sqrt{2}r} $
$ \therefore \frac{B _1}{B _2}=\frac{\sqrt{2}}{1} $
