A proton and electron traveling along with parallel paths entre a region of uniform MF acting perpendicular to their paths.

Question

A proton and electron traveling along with parallel paths entre a region of uniform MF acting perpendicular to their paths. Which of them will move in a circular path with higher frequency.

Answer 1

Centripetal Force for circular motion is:

 

mv²/r = mv(v/r) = mvω

• m = mass
• v = velocity
• r = radius
• ω = angular frequency (= v/r)

 

The force on a particle moving in a constant magnetic field B is:

 

F = qvBsin(θ)

• q = charge
• v - velocity
• B = magnetic field
• θ = angle between the field B and the particles velocity vector

 

θ is given as 90°, so sinθ=1 and we get F = qvB.  Now equate the centripetal and magnetic forces on the particles and solve for the angular frequency, ω:

 

mvω = qvB

ω = qB/m

 

The charge q and the magnetic field B are the same for both the electron and proton.  The proton is about 1800 times more massive than the electron (m_p = 1837·m_e).  Since ω is inversely proportional to mass, the smaller mass (electron) will have the larger frequency (1837 times higher!).