Question

Find ampere force acting on the frame.

Answer 1

Hint: Calculate the both forces considering them as the separate cases of forces between two parallel wires by the formula of the forces between two parallel wires . Both forces act in opposite directions . So , net force will be the resultant of the two forces( forces on two different arms of the same frame) .

Complete step by step solution:


This is the case of force between two parallel current carrying wires. A current carrying wire has an electric as well as magnetic field of its own . Because of this it exerts a force on an object within its electric or magnetic field .
Force on arm 1 because of wire:
$
 \dfrac{{{F_1}}}{a} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2{I_0}I}}{{(na - \dfrac{a}{2})}} \\
{F_{net}} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2{I_0}Ia}}{{(na - \dfrac{a}{2})}} \\ $ ……..(i)
Force on arm 2 and the wire will be :
$
\dfrac{{{F_2}}}{a} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2{I_0}I}}{{(na + \dfrac{a}{2})}} \\
{F_2} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2{I_0}Ia}}{{(na + \dfrac{a}{2})}} \\
 $ …….(ii)
Net force acting on the frame will be the resultant of the two forces :
From eq(i) and (ii):
$
  {F_{net}} = {F_1} - {F_2}
$
$ {F_{net}} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2{I_0}Ia}}{{(n - \dfrac{1}{2})a}} - \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2{I_0}Ia}}{{(n + \dfrac{1}{2})a}} \\
\Rightarrow {F_{net}} = \dfrac{{{\mu _0}}}{{4\pi }} \times 2{I_0}I(\dfrac{1}{{(n - \dfrac{1}{2})}} - \dfrac{1}{{(n + \dfrac{1}{2})}}) \\
\Rightarrow {F_{net}} = \dfrac{{{\mu _0}2{I_0}I}}{{4\pi }}(\dfrac{1}{{{n^2} - \dfrac{1}{4}}}) \\
\Rightarrow {F_{net}} = \dfrac{{{\mu _0}2{I_0}I}}{{4\pi }}(\dfrac{4}{{4{n^2} - 1}}) \\
\Rightarrow {F_{net}} = \dfrac{{2{\mu _0}{I_0}I}}{{\pi (4{n^2} - 1)}} \\
$
We conclude that this is the required answer.

Note: We have to be careful while writing the distances. Both arms of the wires are at different lengths from the wire and hence the distances will be different. If the distance were to be the same , the net force would have been zero and the frame wouldn’t have existed.