The Definition of B — Question 7

Section 28-2 · Physics pages 4504–4505

Question 7 A curved trajectory can have nonzero magnetic force and exactly zero magnetic work A nonrelativistic particle of mass m and charge q follows an arbitrary path in a stationary magnetic field, with no other force. Its initial speed is v0.⃗ vi⃗ vf Schematic; labels, not drawing lengths, specify the data. (a) Prove that its kinetic energy and speed remain constant even if the field varies with position. (b) If its velocity changes from v0ˆx to v0ˆy, find the momentum-change magnitude and explain why this does not contradict zero work.
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Question 7 - Solution A curved trajectory can have nonzero magnetic force and exactly zero magnetic work⃗ vi⃗ vf (a) Method and calculation dK dt = m˙⃗v ·⃗v = q(⃗v ×⃗B) ·⃗v = 0. K = 1 2mv2 0, |⃗v| = v0, WB = ∫⃗ FB · d⃗r = 0. (b) Interpretation and further calculation ∆⃗p = mv0(ˆy −ˆx), |∆⃗p| = √ 2mv0. ∆K = p2 f −p2 i 2m = 0. Physical check Work depends on changes in momentum magnitude squared, whereas impulse depends on the full vector change.

Question 7

A curved trajectory can have nonzero magnetic force and exactly zero magnetic work

A nonrelativistic particle of mass m and charge q follows an arbitrary path in a stationary magnetic field, with no other force. Its initial speed is v0.

Diagram shown in the original SVG worksheet.

  1. Prove that its kinetic energy and speed remain constant even if the field varies with position.

  2. If its velocity changes from v0x̂v_0\hat x to v0ŷv_0\hat y, find the momentum-change magnitude and explain why this does not contradict zero work.

Question 7 - Solution

A curved trajectory can have nonzero magnetic force and exactly zero magnetic work

Diagram shown in the original SVG worksheet.

(a) Method and calculation

dKdt=mv→̇⋅v→=q(v→×B→)⋅v→=0.\frac{dK}{dt}=m\dot{\vec v}\cdot\vec v=q(\vec v\times\vec B)\cdot\vec v=0.K=12mv02,|v→|=v0,WB=∫F→B⋅dr→=0.K=\tfrac12mv_0^2,\quad |\vec v|=v_0,\quad W_B=\int\vec F_B\cdot d\vec r=0.

(b) Interpretation and further calculation

Δp→=mv0(ŷ−x̂),|Δp→|=2mv0.\Delta\vec p=mv_0(\hat y-\hat x),\quad|\Delta\vec p|=\sqrt2mv_0.ΔK=pf2−pi22m=0.\Delta K=\frac{p_f^2-p_i^2}{2m}=0.

Physical check

Work depends on changes in momentum magnitude squared, whereas impulse depends on the full vector change.