KCET 2021 Physics Solved Paper

Given,
T=10‌cm=10×10−2m
I1=I2=10A
Force exerted by one wire on the other per unit length is given as

F

l

=

µ0I1I2

2πr

Substituting the given values, in the above relation, we get

F

l

=

4π×10−7×10×10

2π×10×10−2

=2×10−4Nm−1
Since, the current is flowing in the same direction, so the force will be attractive in nature.