$A$ straight wire of length $0.5 \ m$ and carrying a current of $1.2 \ A$ is placed in a uniform magnetic field of induction $2 \ T$. The magnetic field is perpendicular to the length of the wire. What is the force acting on the wire (in $N$)? $[\sin 90^{\circ} = 1]$

$A$ current of $1 \ A$ is flowing along the positive $x$-axis through a straight wire of length $0.5 \ m$ placed in a region of a magnetic field given by $\vec{B} = (2\hat{i} + 4\hat{j}) \ T$. The magnitude and the direction of the force experienced by the wire are, respectively:

$A$ thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction,as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper,the wire takes the shape of a circle. The tension in the wire is

$A$ semicircular ring of radius $R$ carrying current $i$ is placed in a uniform magnetic field of intensity $B$ such that the plane of the wire is perpendicular to the magnetic field,as shown in the figure. The net force acting on the ring is

If the distance between two current-carrying parallel wires is made $\left(\frac{1}{3}\right)^{rd}$ of its original value,then the magnitude of the force between them (the same current is passing through both wires) is:

Two long straight parallel conductors separated by a distance of $0.5\,m$ carry currents of $5\,A$ and $8\,A$ in the same direction. The force per unit length experienced by each other is