Two long parallel wires each carry 2.2 A in the same direction, with their centers 1.8 cm apart. A. Find the magnitude of the magnetic field halfway between the wires. B. Find the magnitude of the magnetic field at a point in the same plane as the wires, 1.8 cm from one wire and 3.6 cm from the other. C. Find the magnitude of the force of interaction (per length) between the wires.

Explanation:

A )

Magnetic field due to a long wire

B = (μ₀ / 4 π ) x( 2i / r )

= 10⁻⁷ x 2 i / r

where i is current , r is distance of point from the wire

Magnetic field at a point in between the wire will be total of magnetic field generated by both the wires

= 10⁻⁷ [ ( 2 x 2.2 / .009 - 2 x 2.2 / .009 ) ]

= 0

The magnetic field acts in opposite direction so they cancel out each other .

B )

At this point their magnetic field will add up

Net magnetic field

=  10⁻⁷ [ ( 2 x 2.2 / .018 + 2 x 2.2 / .036 ) ]

=  10⁻⁷ [ ( 244.44 + 122.22 ) ]

= 10⁻⁷ x 366.66

= 366.66  x 10⁻⁷ T .

C ) Force on a wire in a magnetic field = BiL where B is magnetic field , i is current and L is length of the wire . If length is one then

force = Bi

magnetic field on each wire

= B = (μ₀ / 4 π ) x( 2i / r )

= 10⁻⁷ x 2 x 2.2 / .018

= 244.44 x 10⁻⁷ T

force on each per unit length

= Bi

= 244.44 x 10⁻⁷ x 2.2

= 537.77 x 10⁻⁷ N /m

This is force of interaction between the two wires . Since direction of current is same , they will attract each other .

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hope this

we need to know the time interval during which the force acts. impulse of force = momentum change