A rectangular loop of wire of size 2.5 cm × 4 cm carries a steady current of 1 A. A straight wire carrying 2 A 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.
(1) `vectau = vecM xx vecB = MBsintheta`
Here M and B have the same direction
θ = 0°
`|vectau| = MBsin theta =0`
(2) We know `vecF_B = i vecI xx vecB`
On line AB, and CD magnetic forces are equal and opposite. So they cancel out each other. Magnetic force on line AD.
`vecF = ivecl xx vecB` (Attractive)
=ilB [l= 4cm = 0.04m]
`because B=(mu_0I) /(2pir)`
`|vecF| = (muiIl)/(2pir)` (Attractive)
Magnetic force on line CB.
`vecF = ivecl xx vecB'` (Repulsive)
`=> F'=|vecF| = ilB'`
`because B' = (mu_0I)/(2pir')`
`F' = (mu_0iIl)/(2pir')` (Repulsive)
So, net force
`F_n = F -F'`
`=(mu_0iIl)/(2pi) [1/r - 1/(r')]`
I = 2A
I = 1 A
r = 2 cm = 0.02 m
r′ = (2 + 2.5) cm = 4.5 cm
= 0.045 m
l = 4 cm = 0.04 m
Plug-in these values in above equation
`F_n = ((4pi xx 10_7)(2)(1)(0.04))/(2pi) [1/0.02 - 1/0.045]`
`= 4 xx 10^-7 xx [1/2 - 1/4.5]`
`= 16 xx 10^-7[0.5 -0.22]`
`= 16 xx 10^-7 [0.278]`
`= 4.45 xx 10^-7 N `
The direction of the force on the loop will be towards left.