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 :
[tex]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[/tex]
The magnetic field due to second wire is given by :
[tex]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[/tex]
Let B is the net magnetic field from wire A, so,
[tex],B=B_1-B_2B=0.006-0.004B = 0.002 T[/tex]
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
