question_answer

A metallic rod of mass per unit length \[0.5\text{ }kg\text{ }{{m}^{1}}\] is lying horizontally on a smooth inclined plane which makes an angle of \[30{}^\circ \] 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 [NEET - 2018]

A)  \[\text{14}\text{.76 A}\]                    

B)  \[\text{5}\text{.98 A}\]

C)  \[\text{7}\text{.14 A}\]                      

D)  \[\text{11}\text{.32 A}\]

Correct Answer: D

Solution :

For equilibrium, \[\text{mg sin30}{}^\circ \text{=I}l\text{B cos 30}{}^\circ \] \[\text{I=}\frac{mg}{l\text{B}}\tan 30{}^\circ \] \[=\frac{0.5\times 9.8}{0.25\times \sqrt{3}}=11.32\,\,\text{A}\]