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:

$A$ massless square loop of wire with resistance $10\,\Omega$ supports a mass of $1\,g$. It hangs vertically with one of its sides in a uniform magnetic field of $10^3\,G$,directed outwards in the shaded region. $A$ $DC$ voltage $V$ is applied to the loop. For what value of $V$ will the magnetic force exactly balance the weight of the supporting mass of $1\,g$? (Given: side length of the loop $= 10\,cm$,$g = 10\,m/s^2$)

$P, Q$ and $R$ are long parallel straight wires in air,carrying currents $20 \ A, 40 \ A$ and $60 \ A$ as shown in the figure. What is the direction of the resultant force on $Q$?

$A$ current-carrying rectangular loop $PQRS$ is made of uniform wire. The lengths are $PR = QS = 5\,cm$ and $PQ = RS = 100\,cm$. If the ammeter current reading changes from $I$ to $2I$,the ratio of magnetic forces per unit length on the wire $PQ$ due to wire $RS$ in the two cases respectively,$f_{PQ}^{I} : f_{PQ}^{2I}$,is:

$A$ circular coil of wire carries a current. $PQ$ is a part of a very long wire carrying a current and passing close to the circular coil. If the directions of currents are those shown in the figure,what is the direction of the force acting on $PQ$?

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: