Question

A current of 6.93 A 6.93 A in a long, straight wire produces a magnetic field of 4.15 μT 4.15 μT at a certain distance from the wire. Find this distance.

Answer 1

Answer:

0.334 m

Explanation:

The magnetic field due to current in a straight thin wire, $B$, is given by

$B = \dfrac{\mu_0 I}{2\pi R}$

where $\mu_0$ is the permeability of free spave with value $4\pi\times10^{-7}$ and $R$ is the distance from the wire.

$R = \dfrac{\mu_0 I}{2\pi B}$

Substituting the values from the question,

$R = \dfrac{4\pi\times10^{-7}\times 6.93}{2\pi\times 4.15\times 10^{-6}}$

$R = \dfrac{2\times10^{-1}\times6.93}{4.15} = 0.334 \text{ m}$

Answer 2

Answer:

0.334m

Explanation:

The magnitude of the magnetic field (B) generated by a current(I) in a wire at a distance r from the wire is given by Biot-Savart law as follows;

B = μ₀ x I / (2 π r)           ---------------------(i)

where;

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

From the question;

B = 4.15μ T = 4.15 x 10⁻⁶ T

I = 6.93A

Substitute these values into equation (i) as follows;

4.15 x 10⁻⁶ = 4π × 10⁻⁷ x 6.93 / (2 π r)

r =  (4π × 10⁻⁷ x 6.93 ) / (4.15 x 10⁻⁶ x 2 π)

r =  (2 × 10⁻⁷ x 6.93 ) / (4.15 x 10⁻⁶)

r =  (2 × 10⁻⁷ x 6.93 ) / (4.15 x 10⁻⁶)

r = 3.34 x 10⁻¹

r = 0.334m

Therefore the distance from the wire is 0.334m