In the figure, a charged particle moves into a region of uniform magnetic field

Question

In the figure, a charged particle moves into a region of uniform magnetic field B, goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 125 ns in the region. (a) What is th decide which). It spends 125 ns in the region. (e) What is the magnitude of B (b) Ir the particle is sent back through the magnetic feld (along the same intial path) but with 3,58 times its previous kinetiç energy, how much time does it spend in the field during this trip? (a) Number Units (b) Number Units

Explanation / Answer

In the magnetic field

magnetic force Fb = q*v*B*sintheta

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

v = speed of proton

B = magnetic field

from newtons second law

Fnet = m*a = m*v^2/r

r = radius of circle

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

magnetic field B = (m/q)*(v/r)

m = masss of proton = 1.67*10^-27 kg

distance travelled d = pi*r = v*t

v/r = pi/t

given t = 125*10^-9 s

magnetic field B = (m/q)*(pi/t) = (1.67*10^-27/(1.6*10^-19))*(pi/(125*10^-9))

magnetic field B = 0.262 T <<<--ANSWER

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part(b)

K2 = 3.85*K1

(1/2)*m*v2^2 = 3.85*(1/2)*m*v1^2

v2 = 1.96*v1

in the magnetic field Fb = Fc

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

radius r = m*v/qB

for speed v1

r1 = m*v1/(q*B)

for speed v2

r2 =m*v2/(qB)

r2/r1 = v2/v1

r2/v2 = r1/v1

for speed v1

time spent

time t1 = pi*r1/v1

for speed v2

time spent

time t2 = pi*r2/v2 = pi*r1/v1 = 125 ns <<----ANSWER

from equation

B = (m/q)*(pi/t)

t is independent of speed v

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