Question

A proton moves with a speed of 1.00 x 106 m/s perpendicular to a magnetic field, B. As a result, the proton moves in circle of radius 20.0 m. If an electron were moving at the same speed perpendicular to the same magnetic field, what would be the radius of the circle the electron would move in

Answer 1

Answer:

0.0109 m ≈ 10.9 mm

Explanation:

proton speed = 1 * 10^6 m/s

radius in which the proton moves = 20 m

determine the radius of the circle in which an electron would move

we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)

For proton :

Mp*V^2 / rp  = qp *VB   ∴  rp = Mp*V / qP*B    ---------- ( 1 )

For electron:

re = Me*V/ qE * B -------- ( 2 )

Next: take the ratio of equations 1 and 2

re / rp = Me / Mp                                 ( note: qE = qP = 1.6 * 10^-19 C )

∴ re ( radius of the electron orbit )

= ( Me / Mp ) rp

= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20

= ( 5.45 * 10^-4 ) * 20

= 0.0109 m ≈ 10.9 mm