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As shown in the figure, a metallic rod of linear density 0.45 \mathrm{~kg} \mathrm{~m}^{-1} is lying horizontally on a smooth inclined 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 \mathrm{~T} magnetic field is acting on it in the vertical upward direction, will be :
{ Use \mathrm{g=10 \mathrm{~m} / \mathrm{s}^{2}}}

Option: 1

30\, \mathrm{A}


Option: 2

15\: \mathrm{A}


Option: 3

10\: \mathrm{A}


Option: 4

3 \: \mathrm{A}


Answers (1)

best_answer

Force on current carrying conductor
is \mathrm{F= BI\ell \sin 90^{\circ}}

\mathrm{F=(0.15) \times i \times \ell \rightarrow(1)}
For stationary rod,

\mathrm{m g \sin \theta=F \cos \theta= B i \ell \cos \theta }

\mathrm{\frac{m g}{\ell}=B i }

\mathrm{0.45 \times 10=0.15 i }

\mathrm{i=30 \mathrm{~A}}

Hence 1 is correct option.






 

Posted by

sudhir.kumar

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