Three straight parallel current-carrying conductors are shown in the figure. The force experienced by the middle conductor of length $25\, cm$ is

$A$ 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:

$A$ current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius $= b$) and $DA$ (radius $= a$) of the loop are joined by two straight wires $AB$ and $CD$. $A$ steady current $I$ is flowing in the loop. The angle made by $AB$ and $CD$ at the origin $O$ is $30^o$. Another straight thin wire with a steady current $I_1$ flowing out of the plane of the paper is kept at the origin. Due to the presence of the current $I_1$ at the origin:

$A$ thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction,as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper,the wire takes the shape of a circle. The tension in the wire is

$A$ vertical wire carrying a current in the upward direction is placed in a horizontal magnetic field directed towards the north. The wire will experience a force directed towards:

The magnetic field existing in a region is given by $\vec B = B_0 \left( 5 + \frac{x}{l} \right) \hat k$. $A$ square loop of edge $l$ and carrying a current $i$ is placed with its edges parallel to $x-y$ axes. Find the magnitude of the net magnetic force experienced by the loop.