AMU Engineering 2017 solved paper Solved Paper

Magnitude of magnetic force per unit length between parallel current carrying wires,
F=

µ0

2π

i1i2

r

Initially, F=

µ0

2π

i1i2

d

As both the wires carry current in same direction, this force is attractive.
Finally i′1=2i and r=3d
So, F′=

µ0

2π

(2i1)i2

3d

=

2

3

F
Since, the direction of current is reversed in one of the wire, hence F′ would be repulsive in nature.
∴F′=−

2

3

F