Question

28. An electron with a speed of 4.0 x 10° m/s enters a uniform magnetic field of magnitude 0.040 T at an angle of 35 degrees to the magnetic field lines. The electron will follow a helical path. a) Determine the radius of the helical path. b) How far forward will the electron have moved after completing one circle?
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Answer 1

Answer:

r= 1.09×10^-4

Explanation:

Given

V=speed=4.0×10^5 m/s

B= magnetic field= 0.040 T

©=angle= 35°

m= mass of electron= 9.11×10^-31

q= charge of electron= 1.60×10^-19

solution

qv×B= mv²/r

qvBsin©=mv²/r

qBsin©=mv/r

r=mv/qBsin©

r=9.11×10^-31× 4.0×10^5/1.064×10^-19×0.04T(sin35°)

r= 1.09×10^-4 m

Answer 2

a) r = 1.09 * $10^{-4}$ m

b) Distance travelled : 6.845 *  $10^{-4}$ m    

What is an electron ?

An electron is a  stable subatomic particle with a charge of negative electricity, found in all atoms and acting as the primary carrier of electricity in solids.

given

charge of electron : 1.6 * $10^{-19}$ C

mass of electron = 9.11 * $10^{-31}$ kg

v = 4.0 x 10 m/s

B =  0.040 T

theta = 35 degrees

since ,

force in magnetic field on electron = centripetal force

a) q(v*B) = m $v^{2}$ / r

q v B sin(theta) =  m $v^{2}$ / r

r =m v /q B sin(theta)

r =  9.11 * $10^{-31}$ * 4.0 x $10^{5}$/ 1.6 * $10^{-19}$ sin (35)

r = 1.09 * $10^{-4}$ m

b)  far forward will the electron have moved after completing one circle will be equal to circumference of the circle = 2πr

 = 2 * 3.14 * 1.09 * $10^{-4}$ m  = 6.845 *  $10^{-4}$ m  

learn more about electron  

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