Question 208 of 276

As shown in the figure, a metallic rod of linear density $0.45kg{m}^{-1}$ 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 $0.15T$ magnetic field is acting on it in the vertical upward direction, will be
(Use $g=10m{s}^{-2}$)

A$30A$B$15A$C$10A$D$3A$

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Answer: (A) $30A$

For the wire to remain stationary, the net force acting on it should be zero.
Balancing the forces along the incline,
$mg\sin 45^{\circ}=ILB\cos 45^{\circ}$
$\Rightarrow I=\left(\dfrac{m}{L}\right)\dfrac{g}{B}$
$\Rightarrow I=\dfrac{\left(0.45\right)\left(10\right)}{0.15}=30A$