🌐Combined Fields - Velocity Selector, Cycloid, Equivalent Gravity

How to use Combined Electric, Magnetic & Gravitational Fields

1. 1On the superposed tab pick a classic case (velocity selector, qE = mg circle, cycloid…) or type q/m, E, B and the launch velocity
2. 2Play the animation to see velocity and force arrows, the drifting guiding centre and the v–t graph, and read the motion type
3. 3On the equivalent-gravity tab set qE/mg, the field direction and the speed to see whether the ball passes the equivalent top
4. 4On the sequential tab choose a mass spectrometer or a multi-stage layout, read the stage analysis, then try the practice problems

FAQ

When does a charged ball move in a uniform circle in combined fields?

When the electric force exactly balances gravity (qE = mg). Only the magnetic force is left; it is always perpendicular to the velocity and supplies the centripetal force, so the ball moves in a uniform circle with r = mv/(qB) and T = 2πm/(qB). If qE and mg do not balance, the ball drifts sideways while it circles.

Why does a particle released from rest in crossed E and B follow a cycloid?

Split the motion into uniform drift at v_d = E/B plus uniform circular motion in the frame moving at v_d; in that frame part of the magnetic force cancels the electric force. Starting from rest, the circular speed is also v_d, so the path is a cycloid. The particle stops once per period, reaches 2mE/(qB²) along E and peaks at 2E/B.

Where is the 'top' of a vertical circle in equivalent gravity?

Combine the constant electric force and gravity into mg′ with g′ = g + qE/m as vectors. The ball's equilibrium lies along g′ (the equivalent bottom), and the opposite point of the circle is the equivalent top. The speed there is smallest when the tension is zero, v = √(g′l), so by work–energy the ball needs at least √(5g′l) at the equivalent bottom.

What is the difference between superposed and sequential fields?

In superposed fields several fields share the same region, so the particle feels all forces at once and you analyse them together. In sequential fields each region holds one field and the particle passes through them in turn: use work–energy or projectile-like motion in electric regions and find the centre, radius and angle in magnetic ones. The key is the speed and direction where one region hands over to the next.

After electric deflection, why is the chord in the magnetic field independent of the deflecting voltage?

If the particle leaves the plates at angle θ, its speed is v = v₀/cosθ and r = mv/(qB) = mv₀/(qB cosθ). With the field edge perpendicular to v₀, the chord is 2r·cosθ = 2mv₀/(qB): the cosθ cancels, so the deflecting voltage drops out. The deflect-then-bend layout shows this result and checks it numerically.