A triangular shaped wire carrying 10A current is placed in a...

Force due to magnetic field on a current carrying conductor is given by, $\overset{\to}{F} = i\overset{\to}{l} \times \overset{\to}{B}$.

The angle between the magnetic field and the direction of the current in the wire $CD$ is $60 °$. Therefore,

$F = ilBsin60 ° ⇒ F = 10 \times 0 . 05 \times 0 . 5 \times \dfrac{\sqrt{3}}{2} ⇒ F = 0 . 216N$