Question

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x

Answer 1

Question:

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.

Answer:

1.6nT [in the negative z direction]

Explanation:

The magnetic field, B, due to a distance of finite value b, is given by;

B = (μ₀IL) / (4πb$\sqrt{b^2 + L^2}$)                -----------(i)

Where;

I = current on the wire

L = length of the wire

μ₀ = magnetic constant = 4π × 10⁻⁷ H/m

From the question,

I = 20A

L = 2.0cm = 0.02m

b = 5.0m

Substitute the necessary values into equation (i)

B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 $\sqrt{5.0^2 + 0.02^2}$)

B = (10⁻⁷ x 20 x 0.02) / (5.0 $\sqrt{5.0^2 + 0.02^2}$)

B = (10⁻⁷ x 20 x 0.02) / (5.0 $\sqrt{25.0004}$)

B = (10⁻⁷ x 20 x 0.02) / (25.0)

B = 1.6 x 10⁻⁹T

B = 1.6nT

Therefore, the magnetic field at the point x = 5.0m  on the x-axis is 1.6nT.

PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.