An electric current I is flowingin a circular wire of radius at. What dose from the centre on the axis of course circular wire will the magnetic field be one eigth of its vaueatthe centre?

The magnetic field at the center of a circular wire carrying a current I is given by the formula:

B = (μ₀ * I) / (2 * π * a)

where:
B = magnetic field strength at the center
μ₀ = permeability of free space (4π x 10^-7 Tm/A)
I = current flowing in the wire
a = radius of the circular wire

To find the distance from the center on the axis of the circular wire where the magnetic field is one-eighth of its value at the center, we can set up the following equation:

B/8 = (μ₀ * I) / (2 * π * x)

Solving for x, we get:

x = a / 2

Therefore, the distance from the center on the axis of the circular wire where the magnetic field is one-eighth of its value at the center is half the radius of the circular wire.

Let us establish a cylindrical coordinate system r, θ, and z with its origin at the center of the circular wire loop, so that the z-axis coincides with the axis of the loop.

If I is the current in the loop, then the magnetic field at a distance r from the center and a distance z along the axis of the loop is given by:

$$B=\dfrac{\mu_0I}{4\pi}\dfrac{2\pi a^2}{(a^2+z^2)^{3/2}}$$

Where a is the radius of the loop and μ0 is the permeability of free space.

We are given that the magnetic field at a distance z from the center is one-eighth of its value at the center. That is :

$$\dfrac{\mu_0I}{4\pi}\dfrac{2\pi a^2}{(a^2+z^2)^{3/2}}=\dfrac{1}{8}\left ( \dfrac{\mu_0I}{4\pi}\dfrac{2\pi a^2}{a^3} \right )$$

Simplifying, we get:

$$(a^2+z^2)^{3/2}=8z^2$$

Taking the cube root of both sides, we get:

$$a^2+z^2=2z^2$$

$$z^2=a^2$$

Therefore, the distance z from the center on the axis of the circular wire where the magnetic field is one-eighth of its value at the center is:

$$z=a$$