In a region of space there is a uniform magnetic field of magnitude 0.25 mT poin

Question

In a region of space there is a uniform magnetic field of magnitude 0.25 mT pointing vertically straight down. (a) If an electron is moving in the horizontal plane at a speed of 550 km s^-1, what will the radius of the resultant circular path be? (b) Will the electron be moving clockwise or counter-clockwise when viewed from above? (c) If a positron (same mass as an electron but with a charge of + q_e) is moving in the horizontal plane at speed of 550 km s^-1, what will the radius of the resultant circular path be? (d) Will the positron be moving clockwise or counter-clockwise when viewed from above? (e) How fast would an alpha particle (^4_2 alpha^2 +) need to be traveling to have a path of the same radius as the electron in part (a)?

Explanation / Answer

part a:

magnetic force will be balanced by the centripetal force.

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

where q=charge

v=speed

B=magnetic field strength

r=radius of the circular path

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

=9.1*10^(-31)*550*1000/(1.6*10^(-19)*0.25*0.001)

=0.012512 m

=1.2512 cm

part b:

as direction of force is determined by product of sign of charge and cross product of initial velocity and magnetic field

so the electron is moving in clockwise direction when viewed from above.

part c:

as mass and charge remains the same, radius will remain the same as electron at 1.2512 cm

part d:

as charge is of opposite sign to that of the electron, the positron will be moving in anticlockwsie direction

part e:

as v=r*q*B/m

==>v=0.012512*2*1.6*10^(-19)*0.25*10^(-3)/(4*1.67*10^(-27))

=149.84 m/s

so speed of the alpha particle required is 149.84 m/s

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