Magnetic Effects Of Current Question 268
Question: A proton (mass $ =1.67\times {{10}^{-27}},kg $ and charge $ =1.6\times {{10}^{-19}},C) $ enters perpendicular to a magnetic field of intensity 2 $ weber/m^{2} $ with a velocity $ 3.4\times 10^{7},m/\sec $ . The acceleration of the proton should be [DPMT 1999]
Options:
A) $ 6.5\times 10^{15},m/{{\sec }^{2}} $
B) $ 6.5\times 10^{13},m/{{\sec }^{2}} $
C) $ 6.5\times 10^{11},m/{{\sec }^{2}} $
D) $ 6.5\times 10^{9},m/{{\sec }^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
F = ma = qvB
Þ $ a=\frac{qvB}{m}=\frac{1.6\times {{10}^{-19}}\times 2\times 3.4\times 10^{7}}{1.67\times {{10}^{-27}}} $ = 6.5 ´ 1015m/sec2
BETA
