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

• Correct answer
  A

  $1 \mathrm{~T}$
• B

  $3 \mathrm{~T}$
• C

  $2.5 \mathrm{~T}$
• D

  $4 \mathrm{~T}$

Solution

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

Hence, the correct answer is option: 'A'.