Question

A particle having a mass of $ 0.5\,g $ carries a charge of $ 2.5\times {{10}^{-8}}C $ . The particle is given an initial horizontal velocity of $ 6\times {{10}^{4}}m{{s}^{-1}} $ . The minimum magnitude of the magnetic field that is required so that, particle will keep moving in a horizontal direction is [take $ g=10\,ms^{-2}$ ]

• A. $ 8.46\,T $
• B. $ 3.33\,T $
• C. $ 8.64\times 10^{-2}T $
• D. None of these

Answer 1

Answer: (B)

To keep the particle moving in horizontal direction $ F_{B} $ should be equal to weight or $ qvB=mg $
[For minimum value of $ B $ angle between $ \mathbf{\vec{v}} $ and $ \vec{\mathbf{B}} $ has to be $ \frac{\pi }{2} $ ]
$ \Rightarrow $ $ B=\frac{0.5\times {{10}^{-3}}\times 10}{2.5\times 10^{-8}\times 6\times 10^{4}}$
$=3.33\,T $