Question

A uniform conducting wire $A B C$, where angle between $A B$ and $B C$ is $37^{\circ}$, has mass of $10\, g$. The wire $A B$ is parallel to $y$ axis. A current of $2 A$ flows through it. If the wire is kept in a uniform magnetic field $B=2 T$ (inwards), then the magnitude acceleration (in $m / s ^{2}$ ) of the wire is (Assume no other force)

Answer 1

Force due to magnetic field
$=i(\vec{l} \times \vec{B})=2 \times[0.03 \hat{i} \times 2(-\hat{k})] $
$=(0.12 \hat{j}) N $
$\therefore $ Acceleration $=\frac{F}{m}$
$=\frac{0.12}{10 \times 10^{-3}}=\left(12\, m / s ^{2}\right) \hat{j}$