A rectangular loop of wire of size 2cm×5cm carries a steady current of 1A. A straight long wire carrying 4A current is kept near the loop as shown in the figure. If the loop and the wire are coplanar, find (i) the 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=μ0I2πr
F=μ0iIl2πr(attractive)
Magnetic force on line CB.
F=|→F|=ilB′ ∵B′=μ0I2πr′
F′=μ0iIl2πr′
This force is repulsive
So net force is
Fn=F−F′=μ0iIl2π(1/r−1/r′)
Now given i=4A, I=1A, r=1cm, r′=(1+2)=3 cm L=5 cm
Putting all the values in the above equation
Fn=2×10−7×4×1×0.05[1/0.01−1/0.03]=26.66×10−7N