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Electrons in Earth’s upper atmosphere have typical speeds near 5.87 ✕ 105 m/s. HINT (a) Calculate the magnitude of Earth’s magnetic field if
Question
Electrons in Earth’s upper atmosphere have typical speeds near 5.87 ✕ 105 m/s. HINT (a) Calculate the magnitude of Earth’s magnetic field if an electron’s velocity is perpendicular to the magnetic field and its circular path has a radius of 6.81 ✕ 10−2 m. T (b) Calculate the number of times per second that an electron circles around a magnetic field line.
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2021-08-30T16:53:42+00:00
2021-08-30T16:53:42+00:00 1 Answers
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Answers ( )
Answer:
a) 4.8×10^-5T
b) 7.38×10^-7seconds
Explanation:
The centripetal force in a magnetic field is given by:
qvB= mv2/r
Where q= charge of the ion
m= mass of the ion
r= radius of the circular path
B= magnetic field
V= velocity
Rearranging to make B subject of formular gives:
B=mv/r
B= [(9.11×10^-31)×(5.87×10^5)]/ [(1.6×10^-19)×(6.81×10^-2)]
B = (5 34×10^-25)/(1.099×10^-20)
B= 4.86×10^-5T
b) T =( 2 pi × r)/v
T = 2 × 3.142 x (6.81×10^-2) / (5.87×10^5
T = 0.428/ (5.87×10^5)
T = 7.38 × 10^-7seconds