Question

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

• A. 1 T
• B. 3T
• C. 2.5 T
• D. 4 T

Answer 1

Answer: (A)

1 T

Answer 2

Here two forces are acting on the rod simultaneously. From the hypothetical free body diagram

mg sin 60 = Bil cos 60$^{\circ}$
$B = \frac{mg}{il} \tan \, 60^{\circ}$
$= \frac{0.01 \times 10}{173 \times 0.1 } \times \sqrt{3}$
$B= 1 T$