$A$ straight current-carrying conductor is kept along the axis of a circular loop carrying current. The force exerted by the straight conductor on the loop is

$A$ straight horizontal conducting rod of length $0.45\; m$ and mass $60\; g$ is suspended by two vertical wires at its ends. $A$ current of $5.0\; A$ is set up in the rod through the wires.
$(a)$ What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
$(b)$ What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) $g = 9.8\; m s^{-2}.$

$A$ single-turn current loop in the shape of a right-angled triangle with sides $5 \,cm, 12 \,cm, 13 \,cm$ carries a current of $2 \,A$. The loop is placed in a uniform magnetic field of magnitude $0.75 \,T$, whose direction is parallel to the current in the $13 \,cm$ side of the loop. The magnitude of the magnetic force on the $5 \,cm$ side is $\frac{x}{130} \,N$. The value of '$x$' is:

$A$ wire carrying a current $i$ is placed in a uniform magnetic field in the form of the curve $y = a \sin \left( \frac{\pi x}{L} \right)$,$0 \leq x \leq 2L$. The force acting on the wire is

Assertion $(A):$ $A$ wire is bent into an irregular shape with the points $P$ and $Q$ fixed. If a current $I$ is passed through the wire,then the area enclosed by the irregular portion of the wire increases.
Reason $(R):$ Wires carrying currents in opposite directions repel each other.

$A$ 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: