Answer:
2.25 x 10⁷ m/s
Explanation:
B = magnitude of magnetic field = 0.032 T
r = radius of the circle = 0.40 cm = 0.0040 m
q = magnitude of charge on the electron = 1.60 x 10⁻¹⁹ C
m = mass of electron = 9.1 x 10⁻³¹ kg
v = speed of the electron
speed of electron is given as
[tex]v = \frac{qBr}{m}[/tex]
[tex]v = \frac{(1.6\times 10^{-19})(0.032)(0.0040)}{9.1\times 10^{-31}}[/tex]
v = 2.25 x 10⁷ m/s
The speed of the electron moving in the given magnetic field is 2.25 x 10⁷ m/s.
The speed of the electron can be determined by equating the forces that act on the electron. These forces include magnetic force and centripetal force.
Fb = Fc
qvB = mv²/r
qB = mv/r
mv = qBr
v = qBr/m
where;
v = (1.6 x 10⁻¹⁹ x 3.2 x 10⁻² x 0.004)/(9.11 x 10⁻³¹)
v = 2.25 x 10⁷ m/s
Thus, the speed of the electron moving in the given magnetic field is 2.25 x 10⁷ m/s.
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