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kolbaska11 [484]
3 years ago
6

3) An electron moves in a circular orbit of radius 0.11 mm, counterclockwise as you look down at it, moving in a plane above and

parallel to the plane of your exam paper. The velocity of the electron is v = 3 × 106m/s. a) Find the magnetic field due to the electron at the center of the circle. b) Is this magnetic field going into the page at the center of the circle or coming towards your face at the center of the circle as you look down at the electron?
Physics
1 answer:
andrey2020 [161]3 years ago
4 0

Answer:

B = 0.15 T

Explanation:

To find the magnitude of the magnetic field you use the following formula:

B=\frac{mv}{qr}   ( 1 )

m: mass of the electron = 9.1*10^-31 kg

v: velocity of the electron = 3*10^6 m/s

q: charge = 1.6*10^-19

r: radius = 0.11mm = 0.11*10^-3 m

You replace the values of the parameters in the equation (1).

B=\frac{(9.1*10^{-31}kg)(3*10^6\frac{m}{s})}{(1.6*10^{-19}C)(0.11*10^{-3}m)}=0.15T

the magnetic field has a magnitude of 0.15 T

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