Question

Question 2 of 7 Attempt 3 A circular coil with 23 turns that has a radius of 24.9 cm carries a current of 2.49 A . What is the strength of the magnetic field at a point on the coil's axis that is 31.9 cm from the center of the coil

Answer 1

6.73 × 10⁻⁵

Step-by-step explanation:

It is given that,

Number of turns, N = 23

Radius of the coil, r = 24.9 cm = 0.249 m

The distance coil axis 31.9 cm, x = 0.319m

Current flowing through the coils, I = 2.49 A

We need to find the magnitude of the magnetic field at a location on the axis of the coils

. The magnetic field of the coils is given by :

`B=(\mu_oINr^2)/((x^2+r^2)^(3/2))B\\=\frac{4\pi * 10^(-7)* 2.49* 23* (0.249)^2}\\{(0.319^2+0.249^2)^(3/2)}B\\ = 6.73 * 10^(-5)\\`