$A$ uniform conducting wire $ABC$ has a mass of $10\,g$. $A$ current of $2\,A$ flows through it. The wire is kept in a uniform magnetic field $B = 2\,T$ directed into the plane of the paper. The acceleration of the wire will be

$A, B$ and $C$ are three parallel conductors of equal lengths carrying currents $I, I$ and $2I$ respectively. The distance between $A$ and $B$ is $x$ and that between $B$ and $C$ is also $x$. $F_1$ is the force exerted by conductor $B$ on $A$. $F_2$ is the force exerted by conductor $C$ on $A$. The current $I$ in $A$ and $I$ in $B$ are in the same direction,and the current $2I$ in $C$ is in the opposite direction. Then:

Two parallel conductors $A$ and $B$ of equal lengths carry currents $I$ and $10\, I$,respectively,in the same direction. Then

Write the magnetic force equation on a current-carrying element $I\vec{dl}$ inside a magnetic field $\vec{B}$. Write the law used to determine the direction of the magnetic force.

$A$ long straight wire carrying a current of $25 \,A$ rests on a table. Another wire $PQ$ of length $1 \,m$ and mass $2.5 \,g$ carries the same current but in the opposite direction. The wire $PQ$ is free to slide up and down. To what height $h$ will wire $PQ$ rise (in $\,mm$)? (Take $g = 9.8 \,m/s^2$ and $\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$)

An infinitely long current-carrying wire and a small current-carrying loop are in the plane of the paper as shown. The radius of the loop is $a$ and the distance of its centre from the wire is $d$ $(d >> a)$. If the loop applies a force $F$ on the wire,then: