Magnetic Effects Of Current Question 144
Question: An electron moves with speed $ 2\times 10^{5} $ m/s along the positive x-direction in the presence of a magnetic induction $ B=\hat{i}+4\hat{j}-3\hat{k} $ (in Tesla.) The magnitude of the force experienced by the electron in Newton’s is (charge on the electron = $ 1.6\times {{10}^{-19}}C) $ [EAMCET 2001]
Options:
A) $ 1.18\times {{10}^{-13}} $
B) $ 1.28\times {{10}^{-13}} $
C) $ 1.6\times {{10}^{-13}} $
D) $ 1.72\times {{10}^{-13}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \overrightarrow{v,}=2\times 10^{5}\hat{i} $ and $ \overrightarrow{B}=(\hat{i}+4\hat{j}-3\hat{k}) $ $ \overrightarrow{F}=q,(\overrightarrow{v,}\times \overrightarrow{B})=-1.6\times {{10}^{-19}}[2\times 10^{5}\hat{i}\times (i+4\hat{j}-3\hat{k})] $ $ =-1.6\times {{10}^{-19}}\times 2\times 10^{5}[\hat{i}\times \hat{i}+4(\hat{i}\times \hat{j})-3(\hat{i}\times \hat{k})] $ $ =-3.2\times {{10}^{-14}}[0+4\hat{k}+3\hat{j}]=3.2\times {{10}^{-14}}(-4\hat{k}-3\hat{k}) $
Þ $ |\overrightarrow{F}|,=3.2\times {{10}^{-14}}\times 5=1.6\times {{10}^{-13}}N. $
BETA
