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weqwewe [10]
3 years ago
9

Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the opposite direction. What is the magnitude of t

he magnetic field halfway between the wires?
Physics
1 answer:
Alex_Xolod [135]3 years ago
5 0

Answer:

The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Explanation:

Given;

distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m

current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively

The magnitude of the magnetic field halfway between the wires can be calculated as;

B = \frac{\mu _oI_1}{2 \pi r} + \frac{\mu_oI_2}{2\pi r}

where;

B is magnitude of the magnetic field halfway between the wires

I₁ is current in the first wire

I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r} (I_1 +I_2)\\\\B = \frac{4\pi *10^{-7} }{2\pi *0.2} (10 +20) = 3.0 *10^{-5}\  T

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

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<u>Given:</u>

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