Q. A circular coil carrying a certain current produces a magnetic field $B_{0}$ at its centre. The coil is now rewound so as to have 3 turns and the same current is passed through it. The new magnetic field at the centre is
A) $\frac{B_{0}}{9}$
B) $9B_{0}$
C) $\frac{B_{0}}{3}$
D) $3B_{0}$
Explanation
(b) The magnetic field produced at the centre of the circular coil carrying current is given by
$B = \frac{\mu_{0}NI}{2r}$
For one turn $N = 1$
$\ \ \ \ B_{0} = \frac{\mu_{0}I}{2r}$
As the coil is rewound
$\ \ \ \ \ r^{'} = \frac{r}{3},\ \ \ \ \ N^{'} = 3$
$\therefore B^{'} = \frac{\mu_{0}I \times 3}{2 \times \left( \frac{r}{3} \right)}$
$\ \ \ \ \ \ \ \ \ = \frac{9\mu_{0}I}{2r} = 9B_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.