Question

A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magnetic field at its center is 6.31Ã10^â4T?

Answer 1

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 =
`6.31* 10^(4) T`

Solution:

To find the number of turns, N, we use the given formula:

`B = (\mu_(o)NI)/(2R)`

Therefore,

`N = (2BR)/(\mu_(o)I)`

`N = (2* 6.31* 10^(4)* 0.06)/(4\pi * 10^(- 7)* 2.65)`

N = 22.74 = 23 turns (approx)