Answer:
A. Attractive
B. ( μ₀I² ) / ( 2πd )
Explanation:
A. We know that currents in the same direction attract, and currents in the opposite direction repel, according to ampere's law. In this case the current in the two wires are flowing in the same direction, and hence the force between the two wires are attractive.
B. Suppose that two wires of length [tex]l_1[/tex] and [tex]l_2[/tex] both carry the current [tex]I[/tex] in the same direction ( given ). In the presence of a magnetic field produced by wire 1, a force of magnitude m say, is experienced by wire 2. The magnitude of the magnetic field produced by wire 1 at distance say d, from it's axis, should thus be the following -
[tex]B_1[/tex] = μ₀I / 2πd
The force experienced by wire 2 should thus be -
[tex]F_2[/tex] = I( [tex]l_2[/tex] [tex]*[/tex] [tex]B_1[/tex] )
= I [tex]*[/tex] [tex]l_2 * B_1 *[/tex] Sin( 90 )
= I [tex]*[/tex] [tex]l_2[/tex] ( μ₀I / 2πd )
Therefore the force per unit length experienced by wire 2 toward wire 1 should be ...
( [tex]F_2[/tex] / [tex]l_2[/tex] ) = ( μ₀I² ) / ( 2πd ) ... which is our solution