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5 01 2019 n-Fall Messages Courses Yardým Logout 5 01 2019 AHMET MERT TURK , (oarenci) PHYS102-Fall Messages Courses Yardym Logout 2018 Ana Menü Contents Grades » FİNot Savisý Evaluate-Feedback Print in lnfo Course Contents .. course contents » An electron is moving at v - 5.75-107 m/s perpendicular to the Earth's magnetic field. If the field strength is 0.521.104 T, what is the radius of the electron's circular path? 6.275 m Computer's answer now shown above. Tries 0/99 This discussion is closed. Send Feedback

Answer 1

Using Force balance on electron moving in circular path:

Fc = Fm

Centripetal Force = Magnetic Force

m*V^2/r = q*V*B

r = m*V/(q*B)

Using given values for electron

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

q = charge of electron = 1.6*10^-19 C

V = 5.75*10^7 m/sec

B = 0.521*10^-4 T

So,

r = 9.1*10^-31*5.75*10^7/(1.6*10^-19*0.521*10^-4)

r = 6.276 m = radius of circular path

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