A proton (q=1.6e-19, m=1.6e-27) is moving perpendicularly inside a magnetic fiel

Question

A proton (q=1.6e-19, m=1.6e-27) is moving perpendicularly inside a magnetic field 500 gauss with energy 2.7 ev (1 ev=1.6e-19 j). Find
Geometry of the path Radius of the path Time period of revolution Frequency of rotation A proton (q=1.6e-19, m=1.6e-27) is moving perpendicularly inside a magnetic field 500 gauss with energy 2.7 ev (1 ev=1.6e-19 j). Find
Geometry of the path Radius of the path Time period of revolution Frequency of rotation
Geometry of the path Radius of the path Time period of revolution Frequency of rotation

Explanation / Answer

particle moves in circular motion

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radius is obtained as from use KE = PE

0.5 mv^2 = eV

so speed v = sqrt(2eV/m)

v = sqrt(2 * 2.7 *1.6 e-19/1.6 e-27)

v = 2.32 *10^4 m/s

so

apply centripetal force = magnetic force

i.e mv^2/r = qvB,

where m = mass o the charged particle

v = velocity,   r = radius ,

q = charge = 1.6*10^-19 C

B = magnetic field

so now , so as r    = mv/qB

r = 1.67 e-27 * 2.32 e 4/(1.6e-19 * 500 e-4)

r = 4.83 mm

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T = 2piR/V

T = (2*3.14* 0.00483)/(2.32 e4)

T = 1.307 us

--------------------------

f = 1/T

f = 1/1.307 e-6

f = 7.64 *10^5 Hz

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