Two long parallel wires are at a distance of $1 \ m$. Both of them carry $1 \ A$ of current. The force of attraction per unit length between the two wires 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:

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 :

The magnetic field existing in a region is given by $\vec B = B_0 [1 + \frac{x}{l}] \hat k$. $A$ square loop of edge $l$ carrying current $I_0$ is placed with its edges parallel to the $x-y$ axes. Find the magnitude of the net magnetic force experienced by the loop.

The horizontal component of the Earth's magnetic field at a certain place is $3 \times 10^{-5} \, T$ and the direction of the field is from geographic south to geographic north. $A$ very long straight conductor is carrying a steady current of $1 \, A$. Calculate the force per unit length on it if the direction of the current is from east to west (in $N/m$).

$A$ straight wire of length $50 \text{ cm}$ carrying a current of $2.5 \text{ A}$ is suspended in mid-air by a uniform magnetic field of $0.5 \text{ T}$. Calculate the mass of the wire. (Take $g = 10 \text{ m s}^{-2}$) (in $\text{ g}$)