In the given figure,the magnetic force on the wire $ABC$ will be $(B = 2 \, T, I = 2 \, A)$.

Two infinitely long parallel conducting wires $A$ and $B$ carry currents $I$ and $2I$, respectively, in the same direction. The wire $A$ has uniform mass per unit length $\lambda$ and lies on an insulated floor. The wire $B$ is kept fixed at a height $h$ above the floor. The minimum magnitude of $h$ so that the wire $A$ does not rise from the floor is : [$g$ is the acceleration due to gravity and $\mu_0$ is the permeability of free space]

Two long straight wires $P$ and $Q$ carrying equal current $10\,A$ each were kept parallel to each other at $5\,cm$ distance. Magnitude of magnetic force experienced by $10\,cm$ length of wire $P$ is $F_1$. If distance between wires is halved and currents on them are doubled,force $F_2$ on $10\,cm$ length of wire $P$ will be :

Two $10 \; cm$ long,straight wires,each carrying a current of $5 \; A$,are kept parallel to each other. If each wire experiences a force of $10^{-5} \; N$,then the separation between the wires is $\dots \; cm$.

$A$ straight wire of length $2 \ m$ carries a current of $10 \ A$. If this wire is placed in a uniform magnetic field of $0.15 \ T$ making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire will be

The figure shows an equilateral triangle $ABC$ of side length '$l$' carrying currents as shown,placed in a uniform magnetic field '$B$' perpendicular to the plane of the triangle. The magnitude of the net magnetic force on the triangle is: