A horizontal rod of mass $10\, gm$ and length $10\, cm$ is placed on a smooth plane inclined at an angle of $60^\circ $ with the horizontal, with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field of induction $B$ is applied vertically downwards. If the current through the rod is $1.73$ $ampere$, then the value of $B$ for which the rod remains stationary on the inclined plane is......$Tesla$
Diffcult
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(c)The given situation can be drawn as follows
$F = ilB$ $ \Rightarrow mg\sin {60^o} = ilB\cos {60^o}$
$ \Rightarrow B = \frac{{0.01 \times 10 \times \sqrt 3 }}{{0.1 \times 1.73}} = 1\;T$
$F = ilB$ $ \Rightarrow mg\sin {60^o} = ilB\cos {60^o}$
$ \Rightarrow B = \frac{{0.01 \times 10 \times \sqrt 3 }}{{0.1 \times 1.73}} = 1\;T$
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