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

Question

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 (A) 1 T (B) 3T (C) 2.5 T (D) 4 T

Tardigrade Admin · Accepted Answer

Here two forces are acting on the rod simultaneously. From the hypothetical free body diagram <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/eff1a530764a991a385785b886914e34-.png" /> mg sin 60 = Bil cos 60° B = (mg/il) tan 60° = (0.01 × 10/173 × 0.1 ) × √3 B= 1 T

1 T

3T

2.5 T

4 T

Here two forces are acting on the rod simultaneously. From the hypothetical free body diagram mg sin 60 = Bil cos 60∘ B=ilmg​tan60∘ =173×0.10.01×10​×3​ B=1T

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

mg sin 60 = Bil cos 60 $^{\circ}$
$B = \frac{m g}{i l} tan 6 0^{\circ}$
$= \frac{0.01 \times 10}{173 \times 0.1} \times 3$
$B = 1 T$