A rectangular loop of wire of size 4cm×10cm carries a steady current of 2A. A straight long wire carrying 5A current is kept near the loop as shown. 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=μiIL2πr
Magnetic force on line CB.
F=|→F|=ilB′ ∴B′=μ0I2πr′
F′=μiIl2πr′
This force is repulsive
So net force is
Fn=F−F′=μ0iIl2π(1/r−1/r′)
Now give i=5A, I=2A , r=1 cm, r'= (1+4)=5cm L= 10 cm
Putting all the values in the above equation
Fn=2×10−7×5×2×0.1[10.01−10.05]=160×10−7N
Due to small distance from the wire attraction is more than repulsion so the loop will move towards the wire.