Lenz's Law — Question 4

Section 30-3 · Physics pages 4798–4799

Question 4 A shrinking loop induces an outward magnetic force on its own boundary A circular resistive loop of radius r is forced to shrink at rate ˙r = −u, where u positive, in a constant out-of-page field B. Resistance R is treated as fixed and self-inductance is negligible. u ⊙B Schematic; labels, not drawing lengths, specify the data. (a) Find induced current and sense. (b) Find the radial force per unit circumference from the applied field and identify whether it aids the shrinkage.
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Question 4 - Solution A shrinking loop induces an outward magnetic force on its own boundary u ⊙B (a) Method and calculation ˙Φ = 2πBr ˙r = −2πBru. i = 2πBru/R > 0 counterclockwise . (b) Interpretation and further calculation d⃗F dℓ= iˆϕ × Bˆz = iBˆr. dF/dℓ= 2πB2ru/R outward . Physical check The magnetic force resists the imposed contraction, consistent with the flux-change rule.

Question 4

A shrinking loop induces an outward magnetic force on its own boundary

A circular resistive loop of radius r is forced to shrink at rate ṙ=−u\dot r=-u, where u positive, in a constant out-of-page field B. Resistance R is treated as fixed and self-inductance is negligible.

Diagram shown in the original SVG worksheet.

  1. Find induced current and sense.

  2. Find the radial force per unit circumference from the applied field and identify whether it aids the shrinkage.

Question 4 - Solution

A shrinking loop induces an outward magnetic force on its own boundary

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Φ̇=2πBrṙ=−2πBru.\dot\Phi=2\pi Br\dot r=-2\pi Bru.i=2πBru/R>0counterclockwise.\boxed{i=2\pi Bru/R>0\quad\text{counterclockwise}}.

(b) Interpretation and further calculation

dF→dℓ=iϕ̂×Bẑ=iBr̂.\frac{d\vec F}{d\ell}=i\hat\phi\times B\hat z=iB\hat r.dF/dℓ=2πB2ru/Routward.\boxed{dF/d\ell=2\pi B^2ru/R\quad\text{outward}}.

Physical check

The magnetic force resists the imposed contraction, consistent with the flux-change rule.