Two parallel wires 10 cm apart carry currents in opposite directions of 8 A. What is the magnetic field halfway between them? (Answer: 6.4 ｘ 10⁻⁵ T)



Given:

Distance between the wire =  d =  10 cm = 0.1 m 

Radius halfway = r = d/2 = 0.1/2 m = 0.05 m

Current through the 1st wire = I₁ = 8 A

Current through the 2ndt wire = I₂ = 8 A

Permeability of free space = µ₀ = 4 ㄫ ｘ 10⁻⁷ H m⁻¹

To Find:

The magnitude of the Magnetic Field halfway between the wires = B = ?


Solution:

The net magnetic Field halfway will be

B = Magnetic field due to 1st wire + Magnetic field due to 2nd wire

B = B₁ + B₂

By using Ampere's Law

B = `\frac {µ₀ I₁}{2ㄫr}`  + `\frac {µ₀ I₂}{2ㄫr}`

As I₂ = I₁ 

So,

B = 2 ｘ `\frac {µ₀ I₁}{2ㄫr}` 

by putting the corresponding values

B = 2 ｘ`\frac { 4ㄫ ｘ 10⁻⁷ H m⁻¹ ｘ8 A}{2ㄫ ｘ 0.05 m}` 

B = 2 ｘ `\frac { 32 ｘ 10⁻⁷ H m⁻¹ A}{0.1 m}`

B =  2 ｘ 320ｘ 10⁻⁷ T

B = 640 ｘ 10⁻⁷ T

B = 6.40 ｘ 10² ｘ 10⁻⁷ T

or 

 B = 6.40 ｘ 10⁻⁵ T ------------------ Ans.


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