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Two long straight parallel wires separated by 1m carry 20A current each in the same direction. Find the magnitude and direction of the magnetic field at a) point midway between the wires; b) point 25cm to the right of point midway between the wires; c) point 75cm to the right of point midway between the wires; d) point 25 cm above one of the wires.

Sagot :

Answer:

Explanation:

a )

At a point midway between the wires , magnetic field due to two wires will be equal and opposite so they will cancel each other . Therefore , net magnetic field will be zero .

b )

At a point 25cm to the right of point midway between the wires

distance of this point will be .25 m from one wire and .75 m from other wire .

Magnetic field due to a current carrying wire can be given by the following expression .

B = (μ₀ / 4π ) x 2I / R where I is current and R is distance of the point

= 10⁻⁷ x 2I / R

For the wire at a distance of .25 m

magnetic field B₁ = 10⁻⁷ x 2 x 20 / .25

= 160 x 10⁻⁷ T .

For the wire at a distance of .75 m

magnetic field B₂ = 10⁻⁷ x 2 x 20 / .75

= 53.33 x 10⁻⁷ T .

These fields are opposite to each other , so

net magnetic field = 160 x 10⁻⁷ T -  53.33 x 10⁻⁷ T

= 106.67 x 10⁻⁷ T .

If the wires are horizontal and current is flowing away from the observer , net magnetic field will be in vertically upward direction .

c )

At a point 75cm to the right of point midway between the wires

The point will be .25 m from one wire and 1 .25 m from other wire . And the magnetic field will be aligned in the same direction .

For the wire at a distance of .25 m

magnetic field B₁ = 10⁻⁷ x 2 x 20 / .25

= 160 x 10⁻⁷ T

For the wire at a distance of 1.25 m

magnetic field B₂ = 10⁻⁷ x 2 x 20 / 1.25

= 32  x 10⁻⁷ T

As the fields are aligned in the same direction

Net magnetic field =  160 x 10⁻⁷ T + = 32  x 10⁻⁷ T

= 192 x 10⁻⁷ T .

If the wires are horizontal and current is flowing away from the observer , net magnetic field will be in vertically downward direction.

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