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Instructions Table of Constants Условие Решение

INPhO 2018, Теория — Задача 5. Magnetic dipoles

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Всички материали от INPhO 2018

Indian National Physics Olympiad – 2018 · 28 януари 2018 г. · 10 т.

Внимание: Бележка към темата

All seven matching archived keys are retained, including alternative ladder methods, unequal-resistor generalization, accepted ranges and the full handwritten graph. Printed refraction, hourglass, Fibonacci and notation inconsistencies remain literal with separate source notes. The full original includes nineteen blank detailed-answer sheets and two graph papers.

Native cover and instructions

Indian National Physics Olympiad – 2018

Date: 28 January 2018. Time: 09:00–12:00 (3 hours). Maximum Marks: 75.

Roll Number: $1\,8\,\square\,\square-\square\,\square\,\square\,\square-\square\,\square\,\square\,\square$ .

I permit/do not permit (strike out one) HBCSE to reveal my academic performance and personal details to a third party.

Besides the International Physics Olympiad (IPhO) 2018, do you also want to be considered for the Asian Physics Olympiad (APhO) 2018? For APhO 2018 and its pre-departure training, your presence will be required in Delhi and Vietnam from April 28 to May 15, 2018. In principle, you can participate in both Olympiads. Yes/No.

Full Name (BLOCK letters) Ms./Mr.: [blank].

Extra sheets attached: [blank box]. Date: [blank]. Centre (e.g. Kochi): [blank]. Signature: [blank].

(Do not write below this line)

Instructions

This booklet consists of 27 pages (excluding this sheet) 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

Quantity	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	$m_e$	$9.11\times10^{-31}\ \mathrm{kg}$
Rest mass of proton	$m_p$	$1.67\times10^{-27}\ \mathrm{kg}$
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	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
Molar mass of water		$18.02\ \mathrm{g\cdot mol^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Question	Marks	Score
1	4
2	8
3	12
4	9
5	10
6	17
7	15
Total	75

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

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

Native key cover and marking policy

Solutions of Indian National Physics Olympiad – 2018

The key cover repeats the date, roll-number form, APhO participation question, nine instructions, thirteen constant rows, seven-question score table and HBCSE address of the question cover, but prints “16 pages (excluding this sheet)”. The retained key has eighteen PDF pages: the cover, sixteen numbered pages and one appended handwritten graph sheet.

On its next page the key states: ANY ALTERNATIVE METHOD OF SOLUTION TO ANY QUESTION THAT IS SCIENTIFICALLY AND MATHEMATICALLY CORRECT, AND LEADS TO THE SAME ANSWER WILL BE ACCEPTED WITH FULL CREDIT. PARTIALLY CORRECT ANSWERS WILL GAIN PARTIAL CREDIT.

Условие

a) Consider two short identical magnets each of mass $M$ and each of which maybe considered as point dipoles of magnetic moment $\vec\mu$ . One of them is fixed to the floor with its magnetic moment pointing upwards and the other one is free and found to float in equilibrium at a height $z$ above the fixed dipole. The magnetic field due to a point dipole at a distance $r$ from it is

\vec B(\vec r)=\frac{\mu_0}{4\pi r^3}\left(3(\vec\mu\cdot\hat r)\hat r-\vec\mu\right)

Obtain an expression for the magnitude of the dipole moment of the magnet in terms of $z$ and related quantities.

Summary answer field: $\vec\mu=$ [blank]. [3 т.]

b) i. [1½ marks] Consider a ring of mass $M_r$ rotating with uniform angular speed about its axis. A charge $q$ is smeared uniformly over it. Relate its angular momentum $\vec S_r$ to its magnetic moment ( $\vec\mu_r$ ).

Summary answer field: $\vec\mu_r=$ [blank].

ii. [1 mark] Assume that the electron is a sphere of uniform charge density rotating about its diameter with constant angular speed. Also assume that the same relation as in the previous part holds between its angular momentum $\vec S$ and its magnetic dipole moment ( $\vec\mu_B$ ). Further assume that $S=h/(2\pi)$ where $h$ is Planck’s constant. Calculate $\mu_B$ .

Summary answer field: $\mu_B=$ [blank].

iii. [3 marks] Assume that the sole contribution to the dipole moment of a $\mathrm{ZnFe_2O_4}$ molecule comes from an unpaired electron. Also assume that the magnets in the part (5a) are $0.482\ \mathrm{kg}$ each of $\mathrm{ZnFe_2O_4}$ and the unpaired electrons of the molecules are all aligned. Calculate the height $z$ . (Note: The molecular weight of $\mathrm{ZnFe_2O_4}=211$ )

Summary answer field: $z=$ [blank].

iv. [1½ marks] In an experiment $\mu_B$ is aligned along a magnetic field of $1\ \mathrm T$ . It is flipped in a direction anti-parallel to the magnetic field by an incident photon. What should be the wavelength of this photon?

Summary answer field: Wavelength = [blank].

Detailed answers can be found on page numbers: [blank]. [7 т.]

Решение

(a)

Given

\vec B(\vec r)=\frac{\mu_0}{4\pi r^3}\left[3(\vec\mu\cdot\hat r)\hat r-\vec\mu\right]

\hat r=\hat k,\quad r=z

\begin{aligned}\vec B(z)&=\frac{\mu_0}{4\pi z^3}[3\mu\hat k-\mu\hat k]\\&=\frac{\mu_0}{4\pi z^3}[2\mu\hat k]\end{aligned}

B(z)=\frac{\mu_0\mu}{2\pi z^3}

The magnets are oppositely facing along $\hat z$ . The force on the upper magnet due to the lower fixed magnet is

\vec F=\frac{3\mu_0\mu^2}{2\pi z^4}

The repulsive magnetic force is balanced by the weight acting down.

Mg=\frac32\frac{\mu_0}{\pi z^4}\mu^2\qquad (16)

\mu^2=\frac23\frac{\pi z^4}{\mu_0}Mg

\mu=\left(\frac{Mg\,4\pi}{6\mu_0}\right)^{1/2}z^2\qquad (17)

(b) i.

Let radius of the ring to be $R$ and it is rotating with the angular speed $\omega\hat z$ . Dipole moment can be expressed as

\vec\mu_r=\text{current}\times\text{area}=\frac{q\omega}{2\pi}\pi R^2\hat z

\vec S_r=M_r\omega R^2\hat z

\Rightarrow\vec\mu_r=\frac q{2M_r}\vec S_r

(b) ii.

\mu_B=\frac{e\hbar}{2m_e}=9.20\times10^{-24}\ \text{to}\ 9.30\times10^{-24}\ \mathrm{A\cdot m^2}

(b) iii.

The system (dipole) is made of $\mathrm{ZnFe_2O_4}$ and has one electron/molecule. The number of molecules is

=\frac{\text{Mass of }\mathrm{ZnFe_2O_4}\text{ in gm}}{\text{Molecular weight of }\mathrm{ZnFe_2O_4}}\times N_A\qquad (18)

=\frac MA\times N_A\qquad (19)

Assuming that each molecule of $\mathrm{ZnFe_2O_4}$ has one unpaired electron and each contributes to dipole moment. Further all of them are perfectly aligned. Hence dipole moment is

\mu=\frac{MN_A\mu_B}A\qquad (20)

Using Eq. (17)

6.60\le z\ (\mathrm{cm})\le6.80

(b) iv.

\text{Wavelength}=\frac{hc}{2\mu_BB}=1.00\ \text{to}\ 1.10\ \mathrm{cm}\qquad (21)

Source note: The key writes a vector $\vec F$ equal to a scalar magnitude, and in Eq. (19) uses $M$ for mass in grams and $A$ for molecular weight without separately defining $A$ ; the question gives the magnet mass in kilograms. The preceding “in gm” wording, formulas and accepted ranges are preserved.

Оригинал в Архива: INPhO2018-Question.pdf · решение от архива: INPhO2018-Solution-20180213.pdf