COMEDK Exam 2013 Solved Paper

The magnetic force between two current carrying wires for current in same direction is attractive. Thus, force on B due to the magnetic field produced due to wire A is
FA=BAIBl=

µ0×IA

2πd

×IB×l
=

0×1×2×l

2πd

=

2µ0l

2πd

, towards A ...........(i)
Similarly, force due to field of wire C,
FC=BCIBl=

µ0ICIBl

2πd

=

µ0×3×2×1

2πd

⇒FC=

6µ0l

2πd

, towards C .............(ii)
From Eqs. (i) and (ii) we get
FC>FA
∴ The resultant force on B is directed towards C.