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:
KE = 10530 J or 10.53 KJ
Explanation:
The formula for kinetic energy is KE = 1/2 mv^2
Let's apply the formula:
KE = 1/2 mv^2
KE = 1/2 (65kg) (18m/s)^2
KE = 10530 J or 10.53 KJ
A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h
There is more wire to travel through,farther distance, and a higher possibility of other disruptions. Please Mark Brainliest!!!
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