A rectangular loop of wire of size 2.5cm × 4cm carries a steady current of 1A. A straight wire carrying 2A current is kept near the loop as shown. If the loop and the wire are coplanar, find the (i) torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire.

(i) →τ=−→M×→B=MBsinθ, here M and B are in the same direction so θ=0 ∴ →τ=0
(ii)We know
FB=i→l×→B
On line AB and CD magnetic forces are equal and opposite. So they cancel out each other.
Magnetic force on line AD is
F=i→l×→B=ilB ∴B=μ0I/2πr
F=μ0iIL/2πr
Magnetic force on line CB.
F'=|→F|=ilB' ∴B' =μ0I/2πr'
F'=μ0iIl/2πr'
This force is repulsive
So net force is
Fn=F−F'=μ0iIl/2π(1/r−1/r')
Now give i=2A, I=1A, r=2cm, r'=(2+2.5)=4.5cm L=4cm
Putting all the values in the above equation
Fn=2×10−7×2×1×0.04[1/0.02−1/0.045]=4.45×10−7N
The direction of the force on the loop will be towards left.