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Natasha_Volkova [10]
1 year ago
13

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

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.
Physics
1 answer:
Ghella [55]1 year ago
5 0

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.

To know more about "angular momentum", refer to the following link:

brainly.com/question/15104254?referrer=searchResults

#SPJ4

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Answer:

Q=977216.256\ J=977.216\ kJ

Explanation:

Given:

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Q=977216.256\ J=977.216\ kJ

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