A metallic rod of mass per unit length 0.5 kg m-1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

Question

A metallic rod of mass per unit length 0.5 kg m-1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is (A) 11.32 A (B) 7.14 A (C) 14.76 A (D) 5.98 A

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7de3bec6af7b371e1bf764baf82fecae-.png" /> For equilibrium, mg sin 30°=I / B cos 30° I =( mg /l B ) tan 30° =(0.5 × 9.8/0.25 × √3) =11.32 A

11.32 A

7.14 A

14.76 A

5.98 A

For equilibrium, mg sin30∘=I/Bcos30∘ I=lBmg​tan30∘ =0.25×3​0.5×9.8​ =11.32A

For equilibrium,
mg $sin 3 0^{\circ} = I / B cos 3 0^{\circ}$
$I = \frac{m g}{lB} tan 3 0^{\circ}$
$= \frac{0.5 \times 9.8}{0.25 \times 3}$
$= 11.32 A$