The resultant force on the current loop $PQRS$ due to a long current-carrying conductor is:

$A$ conductor $ABCDE$,shaped as shown,carries a current $i$. It is placed in the $xy$ plane with the ends $A$ and $E$ on the $x$-axis. $A$ uniform magnetic field of magnitude $B$ exists in the region. The force acting on it will be

$A$ 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:

In the given figure,$X$ and $Y$ are two long straight parallel conductors each carrying a current of $2\,\text{A}$. The force on each conductor is $F$ newtons. When the current in each is changed to $1\,\text{A}$ and reversed in direction,the force on each is now

$A$ horizontal rod of mass $10 \, g$ and length $10 \, cm$ is placed on a smooth plane inclined at an angle of $60^\circ$ with the horizontal,with the length of the rod parallel to the edge of the inclined plane. $A$ uniform magnetic field of induction $B$ is applied vertically downwards. If the current through the rod is $1.73 \, A$,then the value of $B$ for which the rod remains stationary on the inclined plane is......$T$

$A$ uniform conducting wire $ABC$ has a mass of $10 \, g$. $A$ current of $2 \, A$ flows through it. The wire is kept in a uniform magnetic field $B = 2 \, T$. The acceleration of the wire will be ............. $m \, s^{-2}$.