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.

Answer :

**Given:**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 l

_{1}and l

_{2}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 l

_{1}is shown by arrow B

_{1}tangent to the field’s path and Magnetic field at P due to l

_{2}is shown by B

_{2}tangent to the field’s path.

By geometry we can calculate the angle between B

_{1}and B

_{2}which turns out to be 60° .

**Formula used:**

By Ampere’s Law for a current carrying wire is

Where,

B is the magnitude of magnetic field,

μ_{0} is the permeability of free space, μ_{0}= 4π × 10^{-7} T mA^{-1}

d is the distance between the current carrying wire and the required point.

Magnetic field due to wire l_{1}:

∴ B_{1} = 1 × 10^{-4} TMagnetic Field due to l

_{2}:

∴ B

_{2}= 1 × 10

^{-4}T

B

_{1}=B

_{2}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

B

_{R}is the resultant magnetic field at P.

∴ B

_{R}= 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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