Question

Two long straight wires are separated by a distance of d = 0.40 m. The currents are I1 = 1.0 A to the right in the upper wire and I2 = 8.0 A to the left in the lower wire. What are the magnitude and direction of the magnetic field at point P, that is a distance d/2 = 0.20 m below the lower wire?

Answer 1

Answer:

Mangetic field, is 1 μT acting upward due to the current in the upward wire.

Explanation:

Given;

distance of separation of the two wires, d = 0.4 m

the current in the upper wire, I₁ = 1.0 A

the current in the lower wire, I₂ = 8.0 A

The magnetic field at point P, that is a distance d/2 = 0.20 m below the lower wire, is due to the current in the upper wire;

$B = \frac{\mu_oI_1}{2\pi d/2} \\\\B = \frac{4\pi *10^{-7}*1}{2\pi *0.2} = 1*10^{-6} \ T = 1 \mu T$

The direction is upward, since the magnetic field is due to the current in the upward wire.

Therefore, the mangetic field, is 1 μT acting upward due to the current in the upward wire.

Answer 2

Answer:

The magnitude of the magnetic field is 2.34x10⁻⁶T and its direction is opposite to each other.

Explanation:

The equation for Ampere´s law for a straight wire 1 is:

$B_{1} =\frac{u_{o}I_{1} }{2\pi d_{1} }$

I₁ = current flows through wire 1 = 1 A

d₁ = distance from wire 1 to point P = 0.4 m

The equation from wire 1 to point P is:

$d_{1} =d+\frac{d}{2} =\frac{3d}{2}$

$B_{1} =\frac{u_{o}I_{1} }{3\pi d_{1} }=\frac{4\pi x10^{-7}*1 }{3\pi *0.4} =3.33x10^{-7} T$

The equation for Ampere´s law for a straight wire 2 is:

$B_{2} =\frac{u_{o}I_{2} }{2\pi d_{2} }$

The equation from wire 2 to point P is:

$d_{2} =\frac{d}{2}$

$B_{2} =\frac{u_{o}I_{2} }{\pi d_{2} }=\frac{4\pi x10^{-7}*8 }{3\pi *0.4} =2.67x10^{-6} T$

The equation for the net magnetic field is:

$B_{net} =B_{2} -B_{1} =2.67x10^{-6} -3.33x10^{-7} =2.34x10^{-6} T$

The direction of the flowing current is opposite to each other.