question_answer

A conducting rod of length \[l\]and mass m is moving down a smooth inclined plane of inclination 6 with constant velocity v. A current i is flowing in the conductor in a direction perpendicular to paper inwards. A vertically upwards magnetic field \[\mathbf{\vec{B}}\] exists in space. Then magnitude of magnetic field \[\mathbf{\vec{B}}\] is

A) \[\frac{mg}{il}\sin \theta \]                        

B) \[\frac{mg}{il}\tan \theta \]       

C)       \[\frac{mg\cos \theta }{il}\]                       

D) \[\frac{mg}{il\,\sin \,\theta }\]

Correct Answer: B

Solution :

Magnetic force \[{{\overrightarrow{F}}_{m}}=ilB\] acts in the direction as shown in given figure. Rod will move downwards with constant velocity, if net force on it is zero. or         \[{{F}_{m}}\cos \theta =mg\sin \theta \] or         \[ilB\cos \theta =mg\sin \theta \] \[\therefore \]       \[B=\left( \frac{mg}{il} \right)\tan \theta \]