$A$ 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:

$A$ 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.

$A$ straight rod of mass $m$ and length $L$ is suspended from two identical springs as shown in the figure. The springs are stretched by a distance of $x_0$ due to the weight of the rod. The circuit has a total resistance $R$. When a magnetic field perpendicular to the plane of the paper is switched on,the springs are observed to extend further by the same distance $x_0$. The magnetic field strength is:

If two streams of protons move parallel to each other in the same direction,then they

$A$ wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\overrightarrow{B} = (2\hat{i} + 3\hat{j} - 4\hat{k}) \text{ T}$. The magnitude of the magnetic force acting on the wire is $..........IL$.

Two parallel very long straight wires carrying current of $5 \text{ A}$ each are kept at a separation of $1 \text{ m}$. If the currents are in the same direction,the force per unit length between them is . . . . . . $\text{N/m}$. $(\mu_0 = 4\pi \times 10^{-7} \text{ SI units})$