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  4. 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° 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: . Use .g=10 m / s 2

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Q. 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 :
$\left\{\right.$ Use $\left.g=10 \,m / s ^2\right\}$Physics Question Image

JEE MainJEE Main 2022System of Particles and Rotational Motion

Solution:

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$mg \sin 45^{\circ}= ILB \cos 45^{\circ}$
$ \therefore I =\left(\frac{ m }{ L }\right) \frac{ g }{ B }$
$ =\frac{(0.45)(10)}{0.15}=30 A$