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INPhO 2024, Теория — Задача 4. Mag-Grav Tussle

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Всички материали от INPhO 2024

Indian National Physics Olympiad (INPhO)-2024 — Question Paper · 4 февруари 2024 г. · 11 т.

Внимание: Бележка към темата

The original question booklet and the complete available matching key are retained. The separately mentioned Answer Booklet is not included in this retained archive source set. Printed formula, graph-label and prose discrepancies in the key are preserved, rather than silently corrected. In key2(d), only the four vertices of face C′F′E′B′ are explicitly listed, although the question asks for all vertices of the second prism.

Original cover and instructions

Indian National Physics Olympiad (INPhO)-2024

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

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

Question Paper

Date: 04 February 2024

Time: 09:00-12:00 (3 hours)

Maximum Marks: 80

Instructions — Roll No.:

This booklet consists of 9 pages and total of 6 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.

Question Number	1	2	3	4	5	6	Total
Maximum Marks	8	18	14	11	18	11	80

Table of Constants

Universal constant of Gravitation	$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}}$
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	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Molar mass of Helium		$4.003\ \mathrm{g\cdot mol^{-1}}$
Boltzmann constant	$k_B$	$1.3806\times10^{-23}\ \mathrm{J\cdot K^{-1}}$
Earth’s radius	$R_E$	$6371\ \mathrm{km}$
Earth’s mass	$M_E$	$5.97\times10^{24}\ \mathrm{kg}$
Moon’s mass	$M_M$	$7.35\times10^{22}\ \mathrm{kg}$
Moon’s radius	$R_M$	$1737\ \mathrm{km}$

Native key notice

The matching key cover repeats the cover and instructions with “23 pages”.

Please note that alternate/equivalent methods and different way of expressing final solutions may exist. A correct method will be suitably awarded.

Условие

A rectangular conducting loop of mass $m$ , width $w$ , length $h$ , and self inductance $L$ is held in the vertical $x$ - $y$ plane with its bottom edge along the $y$ -axis (see figure on the left below). In this problem take the resistance of the loop to be zero. A uniform magnetic field $\vec B$ is applied horizontally as shown in the figure such that

\begin{aligned}\vec B&=B\hat k&&\text{for }x\le0\\&=0&&\text{for }x>0\end{aligned}

The loop is released from rest at time $t=0$ and descends under gravity (see the figure to the right below). The acceleration due to gravity $g$ is in $+x$ direction.

Bothnativeinitialanddescendingloopdiagramswithmagneticregion,x/yaxes,w/h/h-x/xbracketsand(a)/(b)labels.

a) Obtain $x(t)$ , the position of the bottom edge of the loop at time $t$ , in terms of relevant variables. [5 т.]

b) Imagine different possible scenarios for the nature of motion of the loop and plot $x(t)$ for each. [6 т.]

Решение

(a) Solution:

m\ddot x=mg-BIw\qquad (4.1)

\phi=Bw(h-x)+LI\qquad (4.2)

-IR=\dot\phi=-Bw\dot x+L\dot I\qquad (4.3)

Since $R=0$ Implies

\dot I=\frac{Bw\dot x}L\qquad (4.4)

Differentiating equation of motion with respect to $t$

m\ddot v=-Bw\dot I\qquad (4.5)

\ddot v=\frac{-B^2w^2\dot x}{mL}\qquad (4.6)

=-\omega_0^2v\qquad (4.7)

where

\omega_0^2=\frac{B^2w^2}{mL}\qquad (4.8)

The solution to $v$ is

v=A\cos\omega_0t+D\sin\omega_0t\qquad (4.9)

\dot v=-A\omega_0\sin\omega_0t+D\omega_0\cos\omega_0t\qquad (4.10)

\ddot v=-A\omega_0^2\cos\omega_0t+D\omega_0^2\sin\omega_0t\qquad (4.11)

Applying boundary conditions at $t=0$ , $\dot v=g$ , $v=0$ which implies that

D=\frac g{\omega_0}\qquad (4.12)

A=0\qquad (4.13)

Hence,

v=\frac g{\omega_0}\sin\omega_0t\qquad (4.14)

x=-\frac g{\omega_0^2}\cos\omega_0t+C\qquad (4.15)

at $t=0$ , $x=0$ , which implies that $C=\frac g{\omega_0^2}$ .

Hence,

x=\frac g{\omega_0^2}(1-\cos\omega_0t)\qquad (4.16)

(b) Solution: We found that

x=\frac g{\omega_0^2}(1-\cos\omega_0t)\qquad (4.17)

The frequency of the oscillation is inversely proportional to $m$ , and the amplitude increases with $m$ . When the loop oscillates, the amplitude is

x_m=2\frac g{\omega_0^2}=2\frac{gmL}{B^2w^2}\qquad (4.18)

We can take three limiting cases, $m>m_0$ , $m=m_0$ , and $m<m_0$ , where $m_0=hB^2w^2/2gL$ .

Original colored x(t) graph with all three lower, equal and higher mass cases, h, time labels and pi/omega0.

For $m<m_0$ , the loop oscillates. For $m=m_0$ , the amplitude of the oscillation is $h$ , and the loop comes out of the magnetic field region and falls under gravity. For $m>m_0$ , the loop will come out of the magnetic field quicker it falls under gravity.

Source note: The original plus sign in the second term of (4.11), the prose “inversely proportional to $m$ ”, and all three original motion regimes are retained as printed.

Rough-work pages

Space for rough work — will NOT be submitted for evaluation

The original question and matching key end with blank rough-work pages (question page9 and key page23).

Оригинал в Архива: INPHO2024-Question.pdf · официални решения: INPHO2024-Solution-20240206.pdf