Magnetic Effects Of Current Question 278
Question: A proton enters a magnetic field of flux density $ 1.5,weber/m^{2} $ with a velocity of $ 2\times 10^{7},m/\sec $ at an angle of $ 30{}^\circ $ with the field. The force on the proton will be [MP PET 1994 ; Pb. PMT 2004]
Options:
A) $ 2.4\times {{10}^{-12}},N $
B) $ 0.24\times {{10}^{-12}},N $
C) $ 24\times {{10}^{-12}},N $
D) $ 0.024\times {{10}^{-12}},N $
Show Answer
Answer:
Correct Answer: A
Solution:
$ F=qvB\sin \theta $ $ =1.6\times {{10}^{-19}}\times 2\times 10^{7}\times 1.5\ \sin \ 30^{o} $ $ =1.6\times {{10}^{-19}}\times 2\times 10^{7}\times 1.5\times \frac{1}{2}=2.4\times {{10}^{-12}}N $
BETA
