Magnetic Effects Of Current Question 147
Question: A horizontal rod of mass 10 gm and length 10 cm is placed on a smooth plane inclined at an angle of $ 60{}^\circ $ with the horizontal, with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field of induction B is applied vertically downwards. If the current through the rod is 1.73 ampere, then the value of B for which the rod remains stationary on the inclined plane is
Options:
A) 1.73 Tesla
B) $ \frac{1}{1.73} $ Tesla
C) 1 Tesla
D) None of the above
Show Answer
Answer:
Correct Answer: C
Solution:
The given situation can be drawn as follows $ F=ilB $
$ \Rightarrow mg\sin 60^{o}=ilB\cos 60^{o} $
$ \Rightarrow B=\frac{0.01\times 10\times \sqrt{3}}{0.1\times 1.73}=1\ T $
BETA
