Question

Two long, parallel wires separated by 5.0 mm each carry a current of 60 A. These two currents are oppositely directed. What is the magnitude of the magnetic field at a point that is between the two wires and 2.0 mm from one of the two wires

Answer 1

Given Information:  

Distance between wires = d = 5 mm = 0.005 m

Current in wires = I = 60 A

Required Information:  

Magnetic field at distance 2 mm between two wires  = B = ?  

Answer:  

B = 0.010 T

Explanation:  

When two parallel wires carrying a current I are separated by a distance d then the magnetic field is given by

B1 = μ₀I/2πd1

B2 = μ₀I/2πd2

Where μ₀= 4πx10⁻⁷ is the permeability of free space

The overall magnitude of magnetic field is

B = B1 + B2

Since the two wires are separated by 0.005 m then

d1 = 0.005 - 0.002 = 0.003 m

d2 = 0.002 m

B1 = 4πx10⁻⁷*60/2π*0.003

B1 = 0.004 T

B2 = 4πx10⁻⁷*60/2π*0.002

B2 = 0.006 T

B = 0.004 + 0.006

B = 0.010 T

Therefore, the magnitude of the magnetic field at a point between two wires and 0.002 m away from one of wire is 0.010 T

Answer 2

Answer:

0.002T

Explanation:

It is given that,

Distance between two parallel wires, d = 5 mm = 0.005

Current in each wire, I = 60 A

Let B is the magnitude of the magnetic field at a point that is between the two wires and 2.0 mm from one of the two wires.

calculating the magnetic field due to first wire is given by :

$B_1=\dfrac{\mu_oI}{2\pi d}B_1=\dfrac{4\pi \times 10^{-7}\times 60}{2\pi \times 0.002} B_1=0.006\ T$

The magnetic field due to second wire is given by :

$B_2=\dfrac{\mu_oI}{2\pi d}\\B_2=\dfrac{4\pi \times 10^{-7}\times 60}{2\pi \times 0.003} \\B_2=0.004\ T$

Let B is the net magnetic field from wire A, so,

$,B=B_1-B_2B=0.006-0.004B = 0.002 T$