$A$ current of $5 \ A$ is flowing in a wire of length $1.5 \ m$. $A$ force of $7.5 \ N$ acts on it when it is placed in a uniform magnetic field of $2 \ T$. The angle between the magnetic field and the direction of the current is......$^o$

$A$ straight wire carrying a current $i_1 \, A$ runs along the axis of a circular loop carrying a current $i_2 \, A$. The force of interaction between the two current-carrying conductors is

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 :

$A$ square loop $ABCD$ of side length $L$,carrying a current $i$,is placed near and coplanar with a long straight conductor $XY$ carrying a current $I$. The distance between the conductor and the side $AB$ is $L/2$. The net force on the loop will be:

Two long conductors separated by a distance $d$ carry currents $I_1$ and $I_2$ in the same direction. They exert a force $F$ on each other. Now,the current in one of them is increased to $2$ times and its direction is reversed. The distance between them is also increased to $3d$. The new value of force between them is:

Two very long straight parallel conductors of negligible cross-section carrying equal electric current of $1 \text{ mA}$ are placed one metre apart in a vacuum. The force produced per metre length is $\qquad \text{ N}$. (Given: $\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}$)