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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Answer:
so, why we have to find here..
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Answer:
6.05 cm
Explanation:
The given equation is
2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)
The initial head velocity V₀ₓ =11 m/s
The final head velocity Vₓ is 0
The accelerationis given by =1000 m/s²
the stopping distance = x-x₀=?
So we can wind the stopping distance by following formula
2 (-1000)(x-x₀)=[
]
x-x₀=6.05*
m
=6.05 cm
Answer:
40 meters. look for the dot above the 20 on the x-axis and follow it over to the left.
Explanation:
