$A$ current of $10 \ A$ is flowing in two straight parallel wires in the same direction. The force of attraction between them is $1 \times 10^{-3} \ N$. If the current is doubled in both the wires,the force will be:

$A$ wire carrying a current $I$ is placed inside a uniform magnetic field $\vec{B} = -B_0 \hat{k}$. The shape of the wire is parabolic and has the equation $y = 2x - x^2$. The force on the wire will be:

$A$ $1\, m$ long wire is placed at right angles to a magnetic field. $A$ force of $1\, kg$ wt. acts on it in a magnetic field of $0.98\, T$. The current flowing through the wire is: (in $, A$)

$A$ current of $1 \ A$ is flowing along the positive $x$-axis through a straight wire of length $0.5 \ m$ placed in a region of a magnetic field given by $\vec{B} = (2\hat{i} + 4\hat{j}) \ T$. The magnitude and the direction of the force experienced by the wire are, respectively:

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 :

$AB$ and $CD$ are smooth parallel rails,separated by a distance $l$,and inclined to the horizontal at an angle $\theta$. $A$ uniform magnetic field of magnitude $B$,directed vertically upwards,exists in the region. $EF$ is a conductor of mass $m$,carrying a current $i$. For $EF$ to be in equilibrium,