The Definition of B — Question 5

Section 28-2 · Physics pages 4500–4501

Question 5 One force measurement determines only the magnetic field perpendicular to the velocity A particle of known nonzero q and nonzero velocity v has measured magnetic force F. No electric or mechanical force contributes. x y⃗ v⃗ F = F ˆz B∥unknown Schematic; labels, not drawing lengths, specify the data. (a) Derive the most general B compatible with the measurement and state its consistency condition. (b) For⃗v = vˆx and⃗F = F ˆz with q positive, give B explicitly and identify the field of minimum magnitude.
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Question 5 - Solution One force measurement determines only the magnetic field perpendicular to the velocity x y⃗ v⃗ F = F ˆz B∥unknown (a) Method and calculation⃗ v ·⃗F = 0,⃗ F ×⃗v = q[v2⃗B −(⃗v ·⃗B)⃗v].⃗ B =⃗F ×⃗v qv2 + λˆv, λ arbitrary. (b) Interpretation and further calculation⃗ B = F qv ˆy + λˆx. B2 = ( F qv )2 + λ2 ⇒Bmin = F qv (λ = 0). Physical check Adding a component parallel to the probe velocity leaves the magnetic force unchanged.

Question 5

One force measurement determines only the magnetic field perpendicular to the velocity

A particle of known nonzero q and nonzero velocity v has measured magnetic force F. No electric or mechanical force contributes.

Diagram shown in the original SVG worksheet.

  1. Derive the most general B compatible with the measurement and state its consistency condition.

  2. For v→=vx̂\vec v=v\hat x and F→=Fẑ\vec F=F\hat z with q positive, give B explicitly and identify the field of minimum magnitude.

Question 5 - Solution

One force measurement determines only the magnetic field perpendicular to the velocity

Diagram shown in the original SVG worksheet.

(a) Method and calculation

v→⋅F→=0,F→×v→=q[v2B→−(v→⋅B→)v→].\vec v\cdot\vec F=0,\quad\vec F\times\vec v=q[v^2\vec B-(\vec v\cdot\vec B)\vec v].B→=F→×v→qv2+λv̂,λ arbitrary.\vec B=\frac{\vec F\times\vec v}{qv^2}+\lambda\hat v,\quad\lambda\text{ arbitrary}.

(b) Interpretation and further calculation

B→=Fqvŷ+λx̂.\vec B=\frac F{qv}\hat y+\lambda\hat x.B2=(Fqv)2+λ2⇒Bmin=Fqv(λ=0).B^2=\left(\frac F{qv}\right)^2+\lambda^2\Rightarrow B_{\min}=\frac F{qv}\quad(\lambda=0).

Physical check

Adding a component parallel to the probe velocity leaves the magnetic force unchanged.