Answer:
Given:
radius of the coil, R = 6 cm = 0.06 m
current in the coil, I = 2.65 A
Magnetic field at the center, B = [tex]6.31\times 10^{4} T[/tex]
Solution:
To find the number of turns, N, we use the given formula:
[tex]B = \frac{\mu_{o}NI}{2R}[/tex]
Therefore,
[tex]N = \frac{2BR}{\mu_{o}I}[/tex]
[tex]N = \frac{2\times 6.31\times 10^{4}\times 0.06}{4\pi \times 10^{- 7}\times 2.65}[/tex]
N = 22.74 = 23 turns (approx)
