Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT . The angular momen

Question

2 years ago

13

tum of the electron about the center of the circle is 4.00×10⁻²⁵ kg. m²/s . Determine (b) the speed of the electron.

1 answer:

Ghella [55] · Answer 1 · 2024-01-16T09:11:42+03:00

The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

The angular momentum(L) of an electron moving in a circular path is given by the formula,

L = mvr ........(i)

We know that the radius of the path of an electron in a magnetic field is

r = mv/qB

Putting this value in equation (i),

L = mv x mv/qB

or L = (mv)^2/qB

Putting the given values in the above equation,

4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

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