Question

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

• A. $1.73Tesla$
• B. $\frac{1}{1.73}Tesla$
• C. $1Tesla$
• D. $Noneoftheabove$

Answer 1

The correct option is C $1Tesla$

The given situation can be drawn as follows


$F=ilB\Rightarrow mgsin{60}^{\circ }=ilBcos{60}^{\circ }$
$\Rightarrow B=\frac{0.01\times 10\times \sqrt{3}}{0.1\times 1.73}=1T$