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INPhO 2019, Теория — Задача 5. Parallel metallic rails

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Всички материали от INPhO 2019

Indian National Physics Olympiad – 2019 · 3 февруари 2019 г. · 9 т.

Внимание: Бележка към темата

The complete retained question booklet includes summary answer boxes, eighteen blank detailed-answer pages and three blank graph sheets. All matching archived HBCSE key content is retained, including both full handwritten graph sheets and accepted alternatives/ranges. Printed exponent, sign, factor, notation and unit inconsistencies remain literal with labelled source notes.

Native cover and instructions

Indian National Physics Olympiad – 2019

Date: 03 February 2019. Time: 09:00–12:00 (3 hours). Maximum Marks: 75.

Roll Number: $1\,9\,\square\,\square-\square\,\square\,\square\,\square-\square\,\square\,\square\,\square$ . Extra sheets attached: [blank box]. Centre (e.g. Kota): [blank line].

(Do not write below this line)

This booklet consists of 29 pages (excluding this page) and total of 7 questions.
This booklet is divided in two parts: Questions with Summary Answer Sheet and Detailed Answer Sheet. Write roll number at the top wherever asked.
The final answer to each sub-question should be neatly written in the box provided below each sub-question in the Questions & Summary Answer Sheet.
You are also required to show your detailed work for each question in a reasonably neat and coherent way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of these pages.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations. Strike out any rough work that you do not want to be considered for evaluation.
Adequate space has been provided in the answersheet for you to write/calculate your answers. In case you need extra space to write, you may request additional blank sheets from the invigilator. Write your roll number on the extra sheets and get them attached to your answersheet and indicate number of extra sheets attached at the top of this page.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Use blue or black pen to write answers. Pencil may be used for diagrams/graphs/sketches.
This entire booklet must be returned at the end of the examination.

Table of Constants

Constant	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Planck’s constant	$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}$
	$\hbar$	$h/2\pi$
Universal constant of Gravitation	$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Rest mass of electron	$m_e$	$9.11\times10^{-31}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00\times10^9\ \mathrm{N\cdot m^2\cdot C^{-2}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Question	Marks	Score
1	9
2	11
3	12
4	7
5	9
6	14
7	13
Total	75

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Native key cover

Solutions of Indian National Physics Olympiad – 2019

The native key cover repeats the nine instructions, twelve constant rows, seven-question score table totalling 75 marks, roll-number form and HBCSE address from the question cover. It prints “20 pages (excluding this page)”; the retained key contains twenty-one PDF pages, including two appended handwritten graph sheets.

Условие

A pair of long parallel metallic rails of negligible resistance and separation $w$ are placed horizontally. A horizontal metal rod (dark thick line) of mass $M$ and resistance $R$ is placed perpendicularly onto the rails at one end (as shown). A uniform magnetic field $B$ exists perpendicular to the plane of the paper (pointing into the page). One end of the rail track is connected to a key ( $K$ ) and a capacitor of capacitance $C$ charged to voltage $V_0$ . At $t=0$ the key is closed. Neglect friction and self inductance of the loop.

Entire native rails, thick rod, capacitor polarity, switch K, field into page, w dimension and full circuit.

a) What is the final speed ( $v_{\mathrm{final}}$ ) attained by the rod?

Summary answer box: $v_{\mathrm{final}}=$ [blank]. [6 т.]

b) Consider the ratio $r$ of the maximum kinetic energy attained by the rod to the energy initially stored in the capacitor. What is $r_{\max}$ , the maximum possible value of $r$ , by appropriately choosing $B$ ?

Summary answer box: $r_{\max}=$ [blank]. [2 т.]

c) Let $M=10.0\ \mathrm{kg}$ , $w=0.10\ \mathrm m$ , $V_0=1.00\times10^4\ \mathrm V$ and a bank of capacitors ensures that $C=1.00\ \mathrm F$ . If $r=r_{\max}$ , calculate the value of $v_{\mathrm{final}}$ .

Summary answer box: $v_{\mathrm{final}}(r=r_{\max})=$ [blank].

Detailed answers can be found on page numbers: [blank]. [1 т.]

Решение

(a)

At $t=0$ , $Q_0=CV_0$ . At any instant, $\dot Q=-I$ , where $Q$ is the charge on capacitor and $I$ is the current in RC circuit. From Newton’s second law and Lorentz force,

m\frac{dv}{dt}=IwB\qquad (5.1)

=-\dot QwB\qquad (5.2)

Induced emf in the circuit is $Bwv$ . From KVL,

\frac QC=IR+Bwv\qquad (5.3)

=-\dot QR+Bwv\qquad (5.4)

Differentiating Eq. (5.4) and using Eq. (5.2)

\ddot Q=-\frac1{RC}\dot Q-\frac{B^2w^2}{mR}\dot Q\qquad (5.5)

=-\dot Q\left[\frac1{RC}+\frac{B^2w^2}{mR}\right]\qquad (5.6)

\ddot Q=-\frac{\dot Q}\tau\quad\text{where}\quad\frac1\tau=\frac1{RC}+\frac{B^2w^2}{mR}\qquad (5.7)

\log\left(\frac{\dot Q}{\dot Q_0}\right)=-\frac t\tau\quad\text{where}\quad\dot Q_0=I(t=0)=-\frac{V_0}R\qquad (5.8)

\dot Q=-\frac{V_0}R e^{-t/\tau}\qquad (5.9)

Combining Eqs. (5.2) and (5.9)

m\frac{dv}{dt}=\frac{V_oBw}R e^{-t/\tau}\qquad (5.10)

\int_0^{v_{\mathrm{final}}}m\,dv=\int_0^\infty\frac{V_0Bw}R e^{-t/\tau}\,dt\qquad (5.11)

mv_{\mathrm{final}}=\left.\frac{BwV_0}R(-\tau)e^{-t/\tau}\right|_0^\infty\qquad (5.12)

=\frac{BwV_0\tau}R\qquad (5.13)

v_{\mathrm{final}}=\frac{BwCV_0}{m+B^2w^2c}\qquad (5.14)

Source note: The native question uses rod mass $M$ ; the native derivation uses $m$ . Printed Eq. (5.8) identifies $\dot Q_0$ with $I(t=0)$ although the preceding text states $\dot Q=-I$ . Eq. (5.10) prints $V_o$ with letter $o$ , and Eq. (5.14) prints lowercase $c$ in its denominator while the capacitance elsewhere is $C$ . All original notation and equalities are retained.

(b)

r=\frac{mv_{\max}^2/2}{CV_0^2/2}\qquad (5.15)

=\frac{x}{(1+x)^2}\quad\text{where}\quad x=\frac{B^2w^2C}m\qquad (5.16)

$r$ has a maximum value of $\frac14$ at $x=1$ .

r_{\max}=1/4\qquad (5.17)

(c)

For $r_{\max}$ ,

\frac{B^2w^2C}m=1\Rightarrow B=31.6\ \mathrm T\qquad (5.18)

v_{\max}=1.58\times10^3\ \mathrm{m/s}\qquad (5.19)

Оригинал в Архива: INPhO2019-Question.pdf · решение от архива: INPhO2019-Solution-20190219.pdf