$A$ magnetic field is in the $+Y$ direction. $A$ current $i$ flows through the wire $PQRSTU$. Each side has a length $L$. What is the net force on the wire?

$A$ straight wire of length $2 \ m$ carries a current of $10 \ A$. If this wire is placed in a uniform magnetic field of $0.15 \ T$ making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire will be

$A$ uniform magnetic field $\vec B = (3\hat i + 4\hat j + \hat k) \, T$ exists in a region of space. $A$ semicircular wire of radius $1 \, m$ carrying a current of $1 \, A$ has its centre at $(2, 2, 0)$ and is placed in the $x-y$ plane as shown in the figure. The force on the semicircular wire is:

In the figure shown,a current $I_1$ is established in the long straight wire $AB$. Another wire $CD$ carrying current $I_2$ is placed in the plane of the paper. The line joining the ends of this wire is perpendicular to the wire $AB$. The force on the wire $CD$ is:

$A$ and $B$ are two conductors carrying a current $i$ in the same direction. $x$ and $y$ are two electron beams moving in the same direction.

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