Standard Cases For Force on Current Carrying Conductors
Category : JEE Main & Advanced
Case 1 : When an arbitrary current carrying loop placed in a magnetic field (\[\bot \]to the plane of loop), each element of loop experiences a magnetic force due to which loop stretches and open into circular loop and tension developed in it's each part.
Case 2 : Equilibrium of a current carrying conductor : When a finite length current carrying wire is kept parallel to another infinite length current carrying wire, it can suspend freely in air as shown below
In both the situations for equilibrium of XY it's downward weight = upward magnetic force i.e. \[\mathbf{mg=}\frac{{{\mathbf{\mu }}_{\mathbf{0}}}}{\mathbf{4\pi }}\mathbf{.}\frac{\mathbf{2}{{\mathbf{i}}_{\mathbf{1}}}{{\mathbf{i}}_{\mathbf{2}}}}{\mathbf{h}}\mathbf{.l}\]
Case 3 : Current carrying spring : If current is passed through a spring, then it will contract because current will flow through all the turns in the same direction.
Case 4 : Tension less strings : In the following figure the value and direction of current through the conductor XY so that strings becomes tensionless?
Strings becomes tensionless if weight of conductor XY balanced by magnetic force \[({{F}_{m}})\].
Hence direction of current is from \[X\to Y\] and in balanced condition \[{{F}_{m}}=mg\] \[\Rightarrow \] \[B\,i\,l=mg\] \[\Rightarrow \] \[i=\frac{mg}{B\,l}\]
Case 5 : Sliding of conducting rod on inclined rails : When a conducting rod slides on conducting rails.
In the following situation conducting rod (X, Y) slides at constant velocity if
\[F\cos \theta =mg\sin \theta \]\[\Rightarrow \]\[B\,i\,l\cos \theta =mg\sin \theta \]\[\Rightarrow \]\[B=\frac{mg}{i\,l}\tan \theta \]
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