Explain magnetic turning without an energy gain

Turning is a change of velocity

Can a force act continuously while speed stays constant? Use 0° and +1 T. The positive charge moves right while its magnetic force points down. In a short interval the force changes the direction of velocity. As velocity turns, the force direction turns too, remaining perpendicular. It is incorrect to hold the initial downward force fixed for the whole motion.

Rotate the velocity direction through the available angles. At each angle, check that the force remains perpendicular. The sweep is a collection of instantaneous cases, not a timed animation or a calculated trajectory. Neither slider is elapsed time.

Check the work, not just the force magnitude

Power delivered to a particle is P = F · v = Fx vx + Fy vy + Fz vz, in watts (joules per second). The dot denotes a dot product: multiply matching components and add. It selects the part of force along the motion. With this lab’s components, the sum is q vy Bz vx − q vx Bz vy = 0. Positive and negative terms cancel; the force need not be zero.

At 30° and +1 T, the positive particle has Fx ≈ 8.010 fN, Fy ≈ −13.874 fN, vx ≈ 86,603 m/s and vy = 50,000 m/s. The x contribution to power is positive and the y contribution negative. Use full values before rounding to see cancellation. Small computer roundoff is not a physical energy gain. The numerical ledger displays the exact model result, zero watts.

Kinetic energy is K = mv²/2, where m is mass in kilograms and v is speed in metres per second. With m = 1.67 × 10⁻²⁷ kg and v = 100,000 m/s, K = 8.35 × 10⁻¹⁸ J. A magnetic force alone does no work and does not change this speed or energy. This is not a claim that every device containing magnets is unable to transfer energy: electric fields, mechanical forces or other interactions may also be present. They are absent from this particle model.

Connect the force to a circular path

If the same uniform field extends throughout the path and velocity is perpendicular to it, magnetic force supplies the inward force required for circular motion. Set |q|v|B| = mv²/r, where r is radius, and solve to obtain r = mv/(|q||B|). Radius is a positive length, so use charge and field magnitudes. Their signs choose which way the path curves, not a negative radius.

At +1 T, the predicted radius is about 1.0424 mm for either case. The positive particle initially moving right bends down; viewed on this page, it circles clockwise. The negative particle initially moving right bends up and circles counterclockwise. Equal radii follow from the deliberately equal masses, speed and charge magnitudes. They are not evidence that a real proton and electron would have equal radii at equal speed.

One turn covers distance 2πr. Dividing by speed gives period T = 2πm/(|q||B|), about 65.50 ns at 1 T. Here T is time in seconds; the unit symbol T beside the field slider means tesla. A nanosecond (ns) is 10⁻⁹ seconds. These are conditional analytic predictions, not a numerical integration or observed tracks.

Make paired predictions

Capture 0° at +0.5, +1 and +2 T. Doubling field strength doubles the force and halves radius and period. The kinetic energy remains unchanged. Reversing field changes curvature direction without changing radius, period or energy. Changing the velocity’s in-plane angle changes the starting direction, not radius.

At zero field, the magnetic force vanishes and the particle continues straight under these assumptions. There is no finite circular radius or completed circular period to report. Do not assign zero radius: that would suggest a tight circle, the opposite behavior. The zero-field ledger therefore states straight motion rather than a circular orbit.

Outside this model, doubling speed at fixed mass, charge and nonzero field doubles the radius and quadruples kinetic energy; the nonrelativistic circular period stays the same. A velocity component parallel to the field would remain unchanged while the perpendicular component turns, producing a helix in a uniform field. The controls do not represent that three-dimensional motion. Radiation losses, collisions and relativistic effects are also omitted; a calculated ideal orbit is not an apparatus guarantee.

Transfer to a different particle

A hypothetical particle has mass 4.0 × 10⁻²⁷ kg, charge magnitude 2.0 × 10⁻¹⁹ C and speed 30,000 m/s perpendicular to a 0.50 T uniform field. Calculate its radius, period and kinetic energy. Predict how each changes if speed doubles while the field remains fixed.

Check the magnetic-motion transfer

The radius is 4.0 × 10⁻²⁷ × 30,000 /(2.0 × 10⁻¹⁹ × 0.50) = 0.0012 m = 1.2 mm. The period is 2π × 4.0 × 10⁻²⁷ /(2.0 × 10⁻¹⁹ × 0.50) ≈ 251 ns. Kinetic energy is 1.8 × 10⁻¹⁸ J. Doubling speed gives radius 2.4 mm and energy 7.2 × 10⁻¹⁸ J, with the same period under the stated nonrelativistic approximation.

In your explanation, distinguish force from power, speed from velocity, and the snapshot diagram from the predicted continued motion. Give a condition under which a measured path would fail to be a circle even though a magnetic field is present.

Reference for circular motion and its assumptions: OpenStax: motion in a magnetic field. Numerical comparisons and transfer activities are original.