Question. Source. Yahoo ! Answers !

A singly charged positive ion has a mass of 2.64 10-26 kg. After being accelerated through a potential difference of 240 V, the ion enters a magnetic field of 0.530 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

Solution.

You have to use the Law of Conservation of Energy

Ep – Electric potential energy ; Ek – Mechanical Kinetic energy

Ep = Ek

Ep = q*U

Ek = m * v^2 / 2

q*U = m * v^2 / 2

So, you could find that velocity. v = SQRT(2*q*U/m)

F – centripetal force equal :

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

So, you could find another equation for the velocity

v = q*B*r/m

Now, You have everything in order to calculate radius.

You 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)

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