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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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Which occupies mrs space, 3m cubed or 100ft cubed
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3 years ago
Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

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Question: How is the energy in the food we eat an example of chemical potential energy?
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