Induction and Energy Transfers — Question 4

Section 30-4 · Physics pages 4818–4819

Question 4 A constant applied force initially builds kinetic energy before heating dominates A rod on resistive rails starts from rest under constant horizontal force F0. Set k = B2ℓ2/R and neglect inductance and friction. R rod v ⊗B ℓ F0 Schematic; labels, not drawing lengths, specify the data. (a) Find speed and the fraction of instantaneous input power converted to heat. (b) Find the time when heating and kinetic-energy growth receive equal power.
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Question 4 - Solution A constant applied force initially builds kinetic energy before heating dominates R rod v ⊗B ℓ F0 (a) Method and calculation m˙v = F0 −kv ⇒v = F0 k (1 −e−t/τ), τ = m/k. PR Pin = kv2 F0v = 1 −e−t/τ (t > 0). (b) Interpretation and further calculation PR = ˙K ⇒PR/Pin = 1/2. t = τ ln 2 . Physical check At t=0 both powers vanish, so the ratio there is interpreted as a right-hand limit.

Question 4

A constant applied force initially builds kinetic energy before heating dominates

A rod on resistive rails starts from rest under constant horizontal force F0. Set k=B2ℓ2/Rk=B^2\ell^2/R and neglect inductance and friction.

Diagram shown in the original SVG worksheet.

  1. Find speed and the fraction of instantaneous input power converted to heat.

  2. Find the time when heating and kinetic-energy growth receive equal power.

Question 4 - Solution

A constant applied force initially builds kinetic energy before heating dominates

Diagram shown in the original SVG worksheet.

(a) Method and calculation

mv̇=F0−kv⇒v=F0k(1−e−t/τ),τ=m/k.m\dot v=F_0-kv\Rightarrow v=\frac{F_0}{k}(1-e^{-t/\tau}),\quad\tau=m/k.PRPin=kv2F0v=1−e−t/τ(t>0).\boxed{\frac{P_R}{P_{\mathrm{in}}}=\frac{kv^2}{F_0v}=1-e^{-t/\tau}}\quad(t>0).

(b) Interpretation and further calculation

PR=K̇⇒PR/Pin=1/2.P_R=\dot K\Rightarrow P_R/P_{\mathrm{in}}=1/2.t=τln⁡2.\boxed{t=\tau\ln2}.

Physical check

At t=0 both powers vanish, so the ratio there is interpreted as a right-hand limit.