An electron moves at a speed of 170 m/s perpendicular to the direction of a unif

Question

An electron moves at a speed of 170 m/s perpendicular to the direction of a uniform magnetic field of 0.7 T. What is the radius of the electron's circular orbit in units of nanometers (1 m = 109 nm)?

A PROTON moves at a speed of 30 m/s perpendicular to the direction of a uniform magnetic field of 3 T. What is the radius of the proton's circular orbit in units of nanometers (1 m = 109 nm)?

An electron moves at a speed of 170 m/s perpendicular to the direction of a uniform magnetic field of 0.7 T. What is the radius of the electron's circular orbit in units of nanometers (1 m 109 nm)? 6.77 (in units of nanometers) Remember that the force due to the magnetic field is given by: F QvB sine In this problem, the magnetic force provides the centripetal force necessary to make the particle travel in a circle Therefore, Fcentripetal mv2/r equals Fmagnet -QvB sine. When these two forces are set equal to each other, the radius r can be solved for as r - mv/QB >Do not forget to convert this radius in meters to nanometers by multiplying by 109.

Explanation / Answer

the radius of circular path is given by

r=mv/qB

a) mass of electron =m= 9.11*10-31 kg

v=170 m/s

magnetic field =B= 0.7 T

q=1.6*10-19 C

r=( 9.11*10-31*170/1.6*10-19*0.7) = 1.38*10-9 m = 1.38 nm

corrcet answre is 1.38 nm

b)mass of proton =m= 1.67*10-27 kg

v=30 m/s

magnetic field =B= 3 T

q=1.6*10-19 C

r=( 1.67*10-27*30/1.6*10-19*3) =10.4*10-8 m = 104 nm

corrcet answre is 104 nm

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