The long, staight wires, each carrying an electric current of 5.0 A, are kept parallel to each other at a separation of 2.5 cm. Find the magnitude of the magnetic force experienced by 10 cm of a wire.

Explanation

Solution: The field at the site of one wire due to the
other is
` B=((mu_0)i)/ 2 pi d) = ((2 xx (10^-7)TmA^-1)(5.0A)/ (2*5 xx 10^-2 m)) = 4*0 xx 10^-5 T.`
The force experienced by 10 cm of this wire due to the
other is
F= i lb
` = (5*0 A) (10 xx 10^-2m )(4*0 xx 10^-5 T)`
` = 2*0 xx 10^-5 N.`