Electro Magnetic Induction And Alternating Currents Question 474
Question: A conducting wire of mass m slides down two smooth conducting bars, set at an angle 0 to the horizintal as shown in Fig. The separation between the bars is $ l $ . The system is located in the magnetic field B, perpendicular to the plane of the sliding wire and bars. The constant velocity of the wire is
Options:
A) $ \frac{mg,R,\sin \theta }{B^{2}l^{2}} $
B) $ \frac{mg,R,\sin \theta }{Bl^{2}} $
C) $ \frac{mg,R,\sin \theta }{B^{2}l^{5}} $
D) $ \frac{mg,R,\sin \theta }{Bl^{4}} $
Show Answer
Answer:
Correct Answer: A
Solution:
-
Component of weight along the inclied plane
$ =mg\sin \theta $
Again, $ F=BI\ell =B\frac{B\ell v}{R}\ell =\frac{B^{2}{{\ell }^{2}}v}{R} $
Now, $ \frac{B^{2}{{\ell }^{2}}v}{R}=mg,\sin \theta $
or $ v=\frac{mgR,\sin ,\theta }{B^{2}{{\ell }^{2}}} $
BETA
