$A$ 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 $\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$)

Through two parallel wires $A$ and $B$,$10 \, A$ and $2 \, A$ of currents are passed respectively in opposite directions. If the wire $A$ is infinitely long and the length of the wire $B$ is $2 \, m$,the force on the conductor $B$,which is situated at a $10 \, cm$ distance from $A$,will be:

Consider the inferences given below in respect of the following current loop of wire kept in a magnetic field $\vec{B}$.
$A.$ The force on the element $AC$ of the wire is $\frac{\sqrt{3}}{2}ILB$ directed into the page.
$B.$ The force on the element $AB$ of the wire is $\frac{\sqrt{3}}{2}ILB$ directed into the page.
$C.$ The total force on the loop $ABCA$ is zero.
Which of the above is/are not true?

$A$ current of $10\, A$ is flowing in a wire of length $1.5\, m$. $A$ force of $15\, N$ acts on it when it is placed in a uniform magnetic field of $2\, T$. The angle between the magnetic field and the direction of the current is.....$^o$

$A$ $100$ turn rectangular coil $ABCD$ (in $xy$-plane) is hung from one arm of a balance (Figure). $A$ mass of $500 \text{ g}$ is added to the other arm to balance the weight of the coil. $A$ current of $4.9 \text{ A}$ passes through the coil and a constant magnetic field of $0.2 \text{ T}$ acting inward (in $xz$-plane) is switched on such that only arm $CD$ of length $1 \text{ cm}$ lies in the field. How much additional mass '$m$' must be added to regain the balance?

Give the expression for the force on a current-carrying conductor in a magnetic field.