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Derive the expression for the force between two parallel, current-carrying conductors.

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#### Solution

- Two long parallel current-carrying conductors separated by a distance V in air.
- I
_{1}, I_{2}– current passing through the conductors A and B in the same direction. - Magnetic field due to current I2 of elemental length dl in A

`vec"B"_1 = (mu_0"I"_1)/(2pi"r") (- hat"i")`

`vec"B"_1 = (- mu_0"I"_1)/(2pi"r") hat"i"` - The direction of magnetic field is perpendicular to the plane of the paper and outwards.
- A force on a small elemental length dl in B, experience magnetic field B
_{1}.

`"d"vec"F" = ("I"_2 vec"d"l xx vec"B"_1)`

`= - "I"_2"dl" (mu_0 "I"_1)/(2pi"r") (hat"k" xx hat"i")`

`"d"vec"F" = (- mu_0 "I"_1"I"_2"dl")/(2pi"r")`

Force per unit length of the conductor.

`vec"F"/l = (- mu_0"I"_1"I"_2)/(2pi"r") hat"j"`

Magnetic field due to current I_{2} of elemental

length dl in A

`vec"B"_2 = (mu_0"I"_2)/(2pi"r") hat"i"`

The direction of a magnetic field is perpendicular to the plane of the paper and outwards.

Force on conductor A is directed towards conductor B.

dF = `("I"_1 "d" vec"l" xx vec"B"_2)`

`= "I"_1"dl" (mu_0 "I"_2)/(2pi"r") (hat"k" xx hat"i")`

`"d"vec"F" = (mu_0 "I"_1"I"_2 "dl")/(2pi"r") hat"j"`

Force per unit length of the conductor A,

`vec"F"/l = (- mu"I"_1"I"_2)/(2pi"r") hat"j"`

Attractive force – direction of electric current is same.

Repulsive force – direction of electric current is opposite.