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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Not sure what your question means but the nearest star is Alpha Centauri which is about 4.2 light years (ly) away. This is roughly 4x10¹³ km away. A billion is 10⁹ so this is 4x10⁴ larger than a billion. I'd say the last one then...
Dalton thought that atoms were indivisible particles, and Thomson's discovery of the electron proved the existence of subatomic particles. ... The positive and negative charges cancel producing a neutral atom. images.tutorvista.com. Later discoveries by Rutherford and others lead to additional revisions to atomic theory.
Answer:
the branch of mechanics concerned with the interaction of electric currents with magnetic fields or with other electric currents.
Explanation:
(a) 1200 rad/s
The angular acceleration of the rotor is given by:

where we have
is the angular acceleration (negative since the rotor is slowing down)
is the final angular speed
is the initial angular speed
t = 10.0 s is the time interval
Solving for
, we find the final angular speed after 10.0 s:

(b) 25 s
We can calculate the time needed for the rotor to come to rest, by using again the same formula:

If we re-arrange it for t, we get:

where here we have
is the initial angular speed
is the final angular speed
is the angular acceleration
Solving the equation,
