Q. When a certain length of wire is turned into one circular loop, the magnetic induction at the centre of coil due to some current flowing is $B_{0}$. If the same wire is turned into three loops to make a circular coil, the magnetic induction at the center of this coil for the same current will be
A) $B_{0}$
B) $9\ B_{0}$
C) 3 $B_{0}$
D) 27 $B_{0}$
Explanation
(b) $B_{0} = \frac{\mu_{0}NI}{2a} = \frac{\mu_{0}I \times 1}{2a} = \frac{\mu_{0}I}{2a}$ for 1 turn
For rewinding the coil in three turns, new radius $a/3$, number of turns $\left( N^{'} \right) = 3$
$\therefore$ New magnetic field $= \frac{\mu_{0}I \times 3}{2 \times (a/3)} = \frac{9\mu_{0}I}{2a} = 9B_{0}$
Short trick : $B^{'} = n^{2}B_{0} = (3)^{2}B_{0} = 9B_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.