A charged particle of mass m = 4.1 times 10^-8 degree kg, moving with constant v

Question

A charged particle of mass m = 4.1 times 10^-8 degree kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 2.1T aligned with the positive z-axis as shown. The particle enters the region at (x, y) = (0.77 m, 0) and leaves the region at (x, y) = 0, 0.77 m a time t = 825 mu s after it entered the region. With what speed v did the particle enter the region containing the magnetic field? m/s What is F_x, the x-component of the force on the particle at a time t_1 = 275 mu s after it entered the region containing the magnetic field. What is F_y, the y-component of the force on the particle at a time t_1 = 275 pest after it entered the region containing the magnetic field. What is q, the charge of the particle? Be sure to include the correct sign. mu C If the velocity of the incident charged particle were doubled, how would B have to change (keeping all other parameters constant) to keep the trajectory of the particle the same? Increase B by a factor of 2 Increase B by less than a factor of 2 Decrease B by less than a factor of 2 Decrease B by a factor of 2 There is no change that can be made to B to keep the trajectory the same.

Explanation / Answer

part 1:

as magnetic field and velocity are perpendicular to each other, path followed by the charged particle is circular in nature.

radius of this circular path=0.77 m

so total distance travelled=length of the quarter circle

=(1/4)*2*pi*radius

=0.5*pi*0.77

=1.2095 m

speed=distance/time

=1.2095/(825*10^(-6))

=1466.1 m/s

in circular motion, magnetic force is balanced by centripetal force.

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

==>q=m*v/(r*B)

=4.1*10^(-8)*1466.1/(0.77*2.1)

==>q=3.7174*10^(-5) C

as initial force is along -ve x axis and cross product of v and B is along +ve x axis, charge needs to be negative

hence with proper sign, q=-3.7174*10^(-5) C

part 2:

as speed is constant, angular speed is constant.

angle travelled in 275 us=(275/825)*90 degrees

=30 degrees

for a circular motion, velocity is tangent to the circular path.

so angle made by the velcoity vector with -ve x axis=90-30=60 degrees

velcoity in vector form=v=1466.1*(-cos(60) i +sin(60) j)

=-733.05 i +1269.68 j

magnetic field=B=2.1 k

magnetic force=q*cross product of v and B

=-3.7174*10^(-5)*cross product of (-733.05 i +1269.68 j) and 2.1 k

=-3.7174*10^(-5)*(2666.3 i +1539.4 j)

=-0.099118 i -0.057226 j

hence Fx=-0.099118 N

part 3:

Fy=-0.057226 N

part 4:

q is -3.7174*10^(-5) C=-37.174 uC

part 5:

radius r=m*v/(q*B)
if v is doubled, in order to keep r constant, B needs to be doubled.

hence first option is correct.

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