As shown in the figure, a metallic rod of linear density $0.45\,kg\,m ^{-1}$ is lying horizontally on a smooth incline plane which makes an angle of $45^{\circ}$ with the horizontal. The minimum current flowing in the rod required to keep it stationary, when $0.15\,T$ magnetic field is acting on it in the vertical upward direction, will be $....A$ $\left\{\right.$ Use $\left.g=10 m / s ^{2}\right\}$
A$30$
B$15$
C$10$
D$3$
JEE MAIN 2022, Medium
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A$30$
a $mg \sin 45^{\circ}= ILB \cos 45^{\circ}$
$\therefore I =\left(\frac{ m }{ L }\right) \frac{ g }{ B }$
$=\frac{(0.45)(10)}{0.15}=30\,A$
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