Particle enters a magnetic field. Question 3.

An electron is accelerated from rest through a potential difference of 200 V and then enters a magnetic field of strength 0.02 T acting at right angles to its path. Calculate the radius of the resulting electron orbit.

Solution.

Based on the Law of Conservation of Energy

Ep – Electric potential energy ; Ek – Mechanical Kinetic energy

Ep = Ek

Ep = q*U ; U=200 v

Ek = m * v^2 / 2

q*U = m * v^2 / 2

Now we could find a velocity. v = SQRT(2*q*U/m)

Lorentz force. It is a magnetic force on a point charge.

F = q * v* B * sin (A)

F – centripetal force equal :

F= ma = m * v^2/r = q * v* B * sin (A), sin (90) = 1

Now, we could find another equation for the velocity

v = q*B*r/m

Finally, we have everything in order to calculate radius.

We also could use this equations and get another equation.

v=SQRT(2*q*U/m)
v = q*B*r/m

So, r= SQRT(2*U*m/(q*B^2) ) = SQRT (2*200*9.1*10^-31/(1.6*10^-19*0.02^2)) = 0.002 m

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