Magnetic Effects Of Current Question 279
Question: If a particle of charge $ {{10}^{-12}},coulomb $ moving along the $ \hat{x}- $ direction with a velocity $ 10^{5},m/s $ experiences a force of $ {{10}^{-10}},newton $ in $ \hat{y}- $ direction due to magnetic field, then the minimum magnetic field is [MP PMT 1994]
Options:
A) $ 6.25\times 10^{3},tesla $ in $ \hat{z}- $ direction
B) $ {{10}^{-15}},tesla $ in $ \hat{z}- $ direction
C) $ 6.25\times {{10}^{-3}},tesla $ in $ \hat{z}- $ direction
D) $ {{10}^{-3}},tesla $ in $ \hat{z}- $ direction
Show Answer
Answer:
Correct Answer: D
Solution:
$ F=qvB\sin \theta $
Þ $ B=\frac{F}{qv\sin \theta } $ $ {B_{\min }}=\frac{F}{qv} $ (when q = 90o)
$ \therefore \ {B_{\min }}=\frac{F}{qv}=\frac{{{10}^{-10}}}{{{10}^{-12}}\times 10^{5}}={{10}^{-3}} $ Tesla in $ U=\frac{B^{2}}{2{\mu_{0}}}, $ -direction.
BETA
