question_answer

An inductor of inductance 2 H is connected across two conducting parallel rails of negligible resistances. A conducting rod of length 20 cm and mass of 320 gm is given a speed of 4 m/s along the rails. If a magnetic field of strength 10T is switched on perpendicular to the plane of rails, then maximum current that develops in the inductor will be (neglect friction and gravity)

A)  0.25 A                  

B)  1.2 A

C)  3.6 A                    

D)  1.6 A

Correct Answer: D

Solution :

According to energy conservation principle \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}L{{i}^{2}}\] \[\Rightarrow \,\,i=\sqrt{\frac{m}{L}}v=\sqrt{\frac{320\times {{10}^{-3}}}{2}}\times 4\] \[=4\times 4\times {{10}^{-1}}=1.6\,\,Amp\]