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A horizontal rod of mass $0.01 \mathrm{~kg}$ and length 10 $\mathrm{cm}$ is placed on a frictionless plane inclined at an angle $60^{\circ}$ with the horizontal and with the length of rod parallel to the edge of the inclined plane. A uniform magnetic field is applied 'Vertically downwards. If the current through the rod is $1.73 \mathrm{~A}$, then the value of magnetic field induction $B$ for which the rod remains stationary on the inclined plane is
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The correct answer is:
$1 \mathrm{~T}$
Here two forces acting on the rod simultaneously.
From FBD, mg sin $60=$ Bil $\cos 60^{\circ}$
$\begin{array}{c}
\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{il}} \tan 60^{\circ} \\
=\frac{0.01 \times 10}{173 \times 0.1} \times \sqrt{3}=1 \mathrm{~T}
\end{array}$
From FBD, mg sin $60=$ Bil $\cos 60^{\circ}$
$\begin{array}{c}
\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{il}} \tan 60^{\circ} \\
=\frac{0.01 \times 10}{173 \times 0.1} \times \sqrt{3}=1 \mathrm{~T}
\end{array}$
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