Currents of $10\, A$ and $2\, A$ are passed through two parallel thin wires $A$ and $B$ respectively in opposite directions. Wire $A$ is infinitely long and the length of wire $B$ is $2\, m$. The force acting on the conductor $B$,which is situated at a distance of $10\, cm$ from $A$,will be:

$A$ 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:

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

The figure shows a cross-section of an infinite metal sheet carrying an electric current along its surface. The current per unit length is $J$. $A$ current-carrying square loop is placed nearby the metal sheet such that the plane of the square is perpendicular to the plane of the sheet. Then:

Three infinite straight wires $A, B$ and $C$ carry currents as shown in the figure. The resultant force on wire $B$ is directed:

$A$ conductor lies along the $z$-axis in the range $-1.5 \le z < 1.5 \ m$ and carries a fixed current of $10.0 \ A$ in the $-\hat{a}_z$ direction (see figure). For a magnetic field $\vec{B} = 3.0 \times 10^{-4} e^{-0.2x} \hat{a}_y \ T$,find the power required to move the conductor at a constant speed to $x = 2.0 \ m, y = 0 \ m$ in $5 \times 10^{-3} \ s$. Assume parallel motion along the $x$-axis.