#### Question

A rectangular loop of wire of size 4 cm × 10 cm carries a steady current of 2 A. A straight long wire carrying 5 A 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.

#### Solution

(1) `vectau = vecM xx vec B = MBsintheta`

Here M and B have the same direction

θ = 0°

`|vectau| = MB sintheta = 0`

(2) We know `vecF_B = ivecl 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 [(I = 10cm),(= 0.1cm):]

`because B =( mu_0I)/(2pir)`

`|vecF| = (mui Il)/(2pir)` (Attractive)

Magnetic force on line CB.

`vecF - ivecl xx vecB` [Repulsive]

`=> F= |vecF| = il B'`

`because B' = (mu_0I)/(2pir')`

` F = (mu_0iIl)/(2pir') ` (Repulsive)

So, net force

`F_n = F - F'`

`= (mu_0iIl)/(2pir) [1/r -1/(r')]`

Given *i* = 5A

I = 2 A

*r *= 1 cm = 0.01 m

r′ = (1 + 4) cm = 5 cm

= 0.05 m

*l *= 10 cm = 0.1 m

Plugging in the values in above equation

`F_n = (2 xx 10^-7) (5)(2)(0.1)[1/0.01 - 1/0.05]`

`= 2 xx 10 xx 10^-7 xx 10 [1/1 - 1/5]`

` = 200 xx 10^-7 [(5-1)/5]`

`= 200 xx 10^-7 xx 4/5`

`=160 xx 10^-7 N`