Question

Consider the situation shown in figure. The straight wire is fixed but the loop can move under magnetic force. The loop will

पर्याय

• remain stationary

• move towards the wire

•  move away from the wire

• rotate about the wire.

Answer 1

move towards the wire 

`vec (F}_(AD) + vec(F}_(BC)= 0`

`vec(F)_(AB) > vec(F)_(CD)`

Force acting on the wire per unit length carrying current i₂ due to the wire carrying current i₁ placed at a distance d is given by

\[F = \frac{\mu_o i_1 i_2}{2\pi d}\]

So, forces per unit length acting on sides AB and CD are as follows:

 

\[F_{AB}    =   \frac{\mu_o i_1 i_2}{2\pi d}        (\text{ Towards  the  wire })\] 

\[ F_{CD}    =   \frac{\mu_o i_1 i_2}{2\pi(d + a)}    (\text{ Away  from  the  wire })\]

Here, F_AB > F_CD  because force is inversly proportional to the distance from the wire and wire AB is closer to the wire carrying current i_1.
The forces per unit length acting on sides BC and DA will be equal and opposite, as they are equally away from the wire carrying current i₁, with current i₂ flowing in the opposite direction.
∴ F_BC = - F_DA

Now,
Net force : 

\[F   =    F_{AB}  +  F_{BC}  +  F_{CD}  +  F_{DA} \] 

\[ \Rightarrow F   =   \frac{\mu_o i_1 i_2}{2\pi d} +  F_{BC}  - \frac{\mu_o i_1 i_2}{2\pi(d + a)} -  F_{BC} \] 

\[ \Rightarrow F   =   \frac{\mu_o i_1 i_2}{2\pi}\left( \frac{1}{d} - \frac{1}{d + a} \right)\] 

\[ \Rightarrow F   =   \frac{\mu_o i_1 i_2 a}{2\pi d(d + a)}\]

(Towards the wire)
Therefore, the loop will move towards the wire.