Question

A rectangular loop of wire of size $4cm\times 10cm$ 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.

Answer 1

(i)

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}=MBsin\theta$, here M and B are in the same direction so $\theta =0$

$\therefore \overrightarrow{\tau }=0$

(ii)

We know $F_B=i\overrightarrow{l}\times \overrightarrow{B}$

On line AB and CD magnetic forces are equal and opposite. So they cancel out each other.

Magnetic force on line AD is

$F=i\overrightarrow{l}\times \overrightarrow{B}$

$=ilB$

$\therefore B=\frac{{\mu }_{0}I}{2\pi r}$

$F=\frac{\mu iIL}{2\pi r}$

Magnetic force on line CB.

$F=|\overrightarrow{F}|=il{B}^{\prime }$

$\because B^{\prime }=\frac{{\mu }_{0}I}{2\pi r^{\prime }}$

$F^{\prime }=\frac{{\mu }_{i}Il}{2\pi r}$ This force is repulsive

So net force is

$F_n=F-F^{\prime }$

$=\frac{m{u}_{0}iIl}{2\pi }$ $(\frac{1}{r}-\frac{1}{r^{\prime }})$

Now give i=5A, I=2A , r=1 cm, r'= (1+4)=5cm L= 10 cm

Putting all the values in the above equation

$F_n=2\times {10}^{-7}\times 5\times 2\times 0.1[\frac{1}{0.1}-\frac{1}{0.5}]=160\times {10}^{-7}N$

Due to small distance from the wire attraction is more than repulsion so the loop will move towards the wire.