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  4. A horizontal rod of mass 0.01 kg and length 10 cm is placed on a frictionless plane inclined at an angle 60° 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 A, then the value of magnetic field induction B for which the rod remains stationary on the inclined plane is

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Q. A horizontal rod of mass $0.01\,kg$ and length$ 10\, 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 $1.73\, A$, then the value of magnetic field induction $B$ for which the rod remains stationary on the inclined plane is

VITEEEVITEEE 2015

Solution:

Here two forces are acting on the rod simultaneously. From the hypothetical free body diagram
image
mg sin 60 = Bil cos 60$^{\circ}$
$B = \frac{mg}{il} \tan \, 60^{\circ}$
$ = \frac{0.01 \times 10}{173 \times 0.1 } \times \sqrt{3} $
$B= 1 T $