Two long and parallel straight wires $A$ and $B$ carrying currents of $8.0 \; A$ and $5.0 \; A$ in the same direction are separated by a distance of $4.0 \; cm$. Estimate the force on a $10 \; cm$ section of wire $A$.

$AB$ and $CD$ are smooth parallel rails,separated by a distance $l$,and inclined to the horizontal at an angle $\theta$. $A$ uniform magnetic field of magnitude $B$,directed vertically upwards,exists in the region. $EF$ is a conductor of mass $m$,carrying a current $i$. For $EF$ to be in equilibrium,

$A$ semi-circular current-carrying wire having radius $R$ is placed in the $x-y$ plane with its centre at the origin $O$. $A$ non-uniform magnetic field $\vec B = \frac{B_o x}{2R} \hat k$ (where $B_o$ is a positive constant) exists in the region. The magnetic force acting on the semi-circular wire will be along:

Three long,straight parallel wires carrying currents are arranged as shown. The wire $C$ which carries a current of $5.0 \text{ A}$ is so placed that it experiences no force. The distance of wire $C$ from wire $D$ is (in $cm$)

$A$ single turn current loop in the shape of a right-angled triangle with sides $5\,cm$,$12\,cm$,and $13\,cm$ 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 $..........$

The force acting on a current-carrying wire joining two fixed points $A$ and $B$ in a uniform magnetic field is: