Magnetic Effects Of Current Question 320
Question: A proton moving with a velocity, $ 2.5\times 10^{7}m/s $ , enters a magnetic field of intensity 2.5T making an angle $ 30^{o} $ with the magnetic field. The force on the proton is [AFMC 2000; CBSE PMT 2000]
Options:
A) $ 3\times {{10}^{-12}}N $
B) $ 5\times {{10}^{-12}}N $
C) $ 6\times {{10}^{-12}} $ N
D) $ 9\times {{10}^{-12}}N $
Show Answer
Answer:
Correct Answer: B
Solution:
$ F=qvB\sin \theta =1.6\times {{10}^{-19}}\times 2.5\times 2.5\times 10^{7}\sin 30^{o} $ $ F=1.6\times {{10}^{-19}}\times 6.25\times 10^{7}\times \frac{1}{2}=5\times {{10}^{-12}}N $
BETA
