A quarter wire $\text{AB}$ of radius $R$, carrying a current $i$, is kept in a uniform magnetic field $\overrightarrow{B}$ as shown in the figure. Determine the magnitude of magnetic force acting on the wire?

The correct option is D $\sqrt{2}iRB$
The net force on a current carrying conductor placed in a magnetic field $B$, is given by

$\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B})$

$\overrightarrow{l}=$ length vector/displacement vector.

$\overrightarrow{l_{AB}}=\sqrt{R^2+R^2}=\sqrt{2}R$

$\therefore F=i(\overrightarrow{l_{AB}}\times \overrightarrow{B})$

$F=i\sqrt{2}RB\sin {90}^{\circ }(\because \theta ={90}^{\circ })$

$F=\sqrt{2}iRB$

Hence, option $\left(D\right)$ is the correct answer.