Question

A horizontal rod of mass 0.01 kg and length 10 cm is placed on a frictionless plane inclined at an angle ${60}^o$ 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

• A. 1 T
• B. 3 T
• C. 2.5 T
• D. 4 T

Answer 1

The correct option is A 1 T

Here two forces acting on the rod simultaneously.
From FBD, mg sin 60 = Bil cos 60${}^o$
$B=\frac{mg}{il}tan{60}^o$
$=\frac{0.01\times 10}{173\times 0.1}\times \sqrt{3}=1T$