Q. 105.0( 5 Votes )
Two parallel wires carry equal currents of 10 A along the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point which is 2.0 cm away from each of these wires.
Current in the equal wires: i= 10 A
Distance between the two wires: a = 2 cm = 0.02 m
Distance between the wire and the point : d = 2 cm = 0.02 m
From the diagram l1 and l2 are the wires coming out of the plane.
These two wires and point P form an equilateral triangle of side 0.02 m.
By right hand rule, Magnetic field at P due to l1 is shown by arrow B1 tangent to the field’s path and Magnetic field at P due to l2 is shown by B2 tangent to the field’s path.
By geometry we can calculate the angle between B1 and B2 which turns out to be 60° .
By Ampere’s Law for a current carrying wire is
B is the magnitude of magnetic field,
μ0 is the permeability of free space, μ0= 4π × 10-7 T mA-1
d is the distance between the current carrying wire and the required point.
Magnetic field due to wire l1:
∴ B1 = 1 × 10-4 T
Magnetic Field due to l2 :
∴ B2 = 1 × 10-4 T
B1=B2 as same magnitude of current is flowing through both the wires and point P is located at same distance from each wires.
Now, angle between and : θ = 60°
Resultant of two vectors is given as
BR is the resultant magnetic field at P.
∴ BR = 1.73 × 10-4 T
Hence, resultant magnetic field at a point 2 cm away from due to two current carrying wires is 1.73 × 10-4 T.
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