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;
[tex]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[/tex]
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.
0.3 × 10^(-6))
The magnitude and direction of the magnetic field at point P, that is a distance d/2 = 0.20 m below the lower wire is;
B = 7.7 x 10^(-6) T directed out of the plane of the page.
We are given;
Distance of separation of the two wires; r = 0.4 m
Current in the upper wire; I₁ = 1 A
Current in the lower wire; I₂ = 8.0 A
Now, since point P is d/2 from below the lower wire, it means;
Distance of upper wire from point P = 0.4 + (0.4/2) = 0.6 m
Distance of lower wire from point P = 0.2 m
Now, the magnitude and direction of the magnetic field at point P is given by;
B_net = B₂ - B₁
Formula for Magnetic field is;
B = (μ_o*I)/(2πr)
Where;
μ_o = 4π x 10^(-7) T.m/A
Thus;
B₁ = (4π × 10^(-7) × 1)/((2π × 0.6)
B₁ = 0.3 × 10^(-6) T
Similarly;
B₂ = (4π × 10^(-7) × 8)/((2π × 0.2)
B₂ = 8 × 10^(-6) T
Thus;
B_net = (8 × 10^(-6)) - (0.3 × 10^(-6))
B_net = 7.7 × 10^(-6) T
The direction will be out of the plane of the page.
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