An electron at point a in the figure (figure 1) has a speed v0 of 1.40×106 m/s .

An electron at point A in (Figure 1) has actually a speed v0 of 1.40×106
m/s . a. get the magnitude of this magnetized industry that’ll trigger
the electron to adhere to the semicircular road from the to B. b. Discover
enough time necessary for the electron to maneuver from the to B. c. Exactly What
magnetized industry is required in the event that particle had been a proton
rather than an electron?

distance of circular road is r = mv/qB
therefore D = 10 cm
roentgen = 5 cm
magnetized industry B = mv/qr
right here semicircle
for electron :
B = (9.11*10^-31 * 1.4*10^6)/(1.6*10^-19 * 5*10^-2)
B = 1.6 *10^-4 Tesla
——————————
Circumference C = 2pi Roentgen
C = 2*3.14* 0.05 = 0.314 m
amount of semi-circle = 2 piR /2 = 0.157 m
therefore
time = s/v
t = 0.157/1.4 age 6
t = 112.1 ns
———————————————
for proton,
B = (1.67*10^-27 * 1.4*10^6)/(1.6*10^-19 * 5*10^-2)
B = 0.29 Tesla
0 0 10.0 cm
An electron at point A in (Figure 1) has a speed v0 of 1.40×106
m/s . a. Find the magnitude of the magnetic field that will cause
the electron to follow the semicircular path from A to B. b. Find
the time required for the electron to move from A to B. c. What
magnetic field would be needed if the particle were a proton
instead of an electron?

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