Physics 2. Please show all steps and equations for the following 2 problems Phys
ID: 2030607 • Letter: P
Question
Physics 2. Please show all steps and equations for the following 2 problems Physics 2. Please show all steps and equations for the following 2 problems Physics 2. Please show all steps and equations for the following 2 problemsm/s in the tx-direction into a mis and charge 10 uC is passes through the origin with a velocity of2 in the-y-direction. (a) If there is no electrical field and the particle continues moving in a s constant speed, what is the magnetic force on the particle? (b) What is the magnitude a the magnetic field? (o) If the particle attained its speed by sliding down a ramp before entering the magnetic field, what was the height of the ramp? orm magnetic field parallel to the z-axis and a uniform gravitational field of 1 g
Explanation / Answer
(a)
magnetic frceFb = q*(v x B )
q = 10*10^-6 C
v = 2 i ( along +x direction )
B = Bz k ( along z direction)
gravitational force Fg = m*g (-j) - y direction
as the charge is moving straight with constant speed acceleration = 0
Fnet = 0
Fb + Fg = 0
Fb = -Fg
Fb is along +y direction
Bz is along -z direction
therefore
Fb = Fg = 2*10^-3*9.8 = 0.0196 N
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(b)
Fb = Fg
q*v*Bz = m*g
10*10^-6*2*Bz = 2*10^-3*9.8
magnitude of magnetic field B = 980 T
direction of magnetic field is along -z direction
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potential energy at the top of ramp = kinetic energy at the bottom
PE = KE
m*g*h = (1/2)*m*v^2
h = v^2/(2g)
h = 2^2/(2*9.8)
h = 0.204 m
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In the magnetic field magnetic force Fb = q*v*B
the force is perependicular to speed
the particle moves in circular path
In circular centripetal force Fc = Fb
m*v^2/r = q*v*B
m*v/r = q*B
r = mv/(qB)
time period T = 2pir/v
T = 2*pi*m*v/(v*qB)
T = 2*pi*m/(qB)
frequency = f = 1/T = qB/(2*pi*m)
given frequency f = 2.4*10^6 Hz
2.4*10^6 = 1.6*10^-19*B/(2*pi*9.11*10^-31)
B = 8.58*10^-5 T <<<-------------ANSWER
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part(b)
speed v = 2*pi*r/T
speed v = 2*pi*r*f = 2*pi*0.4*2.4*10^6 = 6.03*10^6 m/s <<--------ANSWER
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part(c)
in the electron gun
work done = change in KE
deltaV*q = (1/2)*m*v^2
volatge delta V = m*v^2/(2*q)
voltage = 9.11*10^-31*(6.03*10^6)^2/(2*1.6*10^-19) = 103.5 V <<<-------ANSWER