$A$ wire loop of irregular shape carrying current is placed in an external magnetic field. If the wire is flexible,the shape of the loop changes to

Two long conductors separated by a distance $d$ carry currents $I_1$ and $I_2$ in the same direction. They exert a force $F$ on each other. Now,the current in one of them is increased to $2$ times and its direction is reversed. The distance between them is also increased to $3d$. The new value of force between them is:

$A$ massless square loop of wire of resistance $R$ supporting a mass $M$ hangs vertically with one of its sides in a uniform magnetic field $B$ directed outwards in the shaded region. $A$ d.c. voltage $V$ is applied to the loop. For what value of $V$ will the magnetic force exactly balance the weight of the supporting mass $M$? (side of loop $= L, g =$ acceleration due to gravity)

$A, B$ and $C$ are three parallel conductors of equal lengths carrying currents $I, I$ and $2I$ respectively. The distance between $A$ and $B$ is $x$ and that between $B$ and $C$ is also $x$. $F_1$ is the force exerted by conductor $B$ on $A$. $F_2$ is the force exerted by conductor $C$ on $A$. The current $I$ in $A$ and $I$ in $B$ are in the same direction,and the current $2I$ in $C$ is in the opposite direction. Then:

$A$ semi-circular current-carrying wire having radius $R$ is placed in the $x-y$ plane with its centre at the origin $O$. $A$ non-uniform magnetic field $\vec B = \frac{B_o x}{2R} \hat k$ (where $B_o$ is a positive constant) exists in the region. The magnetic force acting on the semi-circular wire will be along:

In the figure,the cube has an edge length of $40\,cm$. Four straight segments of wire $ab$,$bc$,$cd$,and $da$ form a closed loop that carries a current $I = 5\,A$. $A$ uniform magnetic field of $0.02\,T$ is in the $+y$-direction. The ratio of the magnetic force on segment $ab$ to the magnetic force on segment $bc$ is: