Two long parallel copper wires carry currents of $5\,A$ each in opposite directions. If the wires are separated by a distance of $0.5\,m$,then the force between the two wires is

If a current is passed through a spring,then the spring will:

$A$ straight wire of length $2 \ m$ carries a current of $10 \ A$. If this wire is placed in a uniform magnetic field of $0.15 \ T$ making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire 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 long straight parallel wires,carrying (adjustable) current $I_1$ and $I_2$,are kept at a distance $d$ apart. If the force $F$ between the two wires is taken as 'positive' when the wires repel each other and 'negative' when the wires attract each other,the graph showing the dependence of $F$ on the product $I_1 I_2$ would be

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by $15 \ cm$ length of wire $Q$ is . . . . . . . $(\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A)$