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INPhO 2026, Теория — Задача 2. Current affairs

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Всички материали от INPhO 2026

Indian National Physics Olympiad (INPhO)-2026 · Question Paper · 1 февруари 2026 г. · 12 т.

Внимание: Бележка към темата

Matching original question and official key booklets are retained in full. The separate blank Answer Booklet, Summary Answersheet copies and blank graph paper referred to in the questions are absent from the archive; the supplied completed Kepler/glass answer-sheet illustrations remain in the key. Native question pages7 and9 are blank. Question6 says photographs on the next page although its photos occur on native p10 after blank p9. Printed source errors/conventions are preserved and attributed locally: P2 negative dM followed by positive boxed dUm and changing absolute-work notation; P3 Qa=-W followed by positive released heat, dropped-mass wording and P0/V0 notation; P4 derivative/height signs, c/cs and time-delay description; P5 intermediate phiP versus final phiC, log log and r/logarithm numerical scales. The graph/grid pixels and printed fitted results are source data, not newly calculated measurements.

Question booklet · Instructions and Table of Constants

Indian National Physics Olympiad (INPhO)-2026

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION · Tata Institute of Fundamental Research · V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Question Paper

Date: 01 February 2026 · Time: 09:00–12:00 (3 hours) · Maximum Marks: 75

Roll Number: □□□□ – □□□□ – □□□□

Instructions

This booklet consists of 10 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.
Take as many data points as possible for the analysis.
Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.
Except for writing your roll number, no rough work or scribbling is allowed on the question paper.

Table of Constants

Quantity	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
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	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Boltzmann constant	$k_B$	$1.3806\times10^{-23}\ \mathrm{J\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$
One Atmospheric pressure	atm	$1.013\times10^5\ \mathrm{Pa}$

Question Number	1	2	3	4	5	6	Total
Maximum Marks	12	12	11	10	18	12	75

Solution booklet · Instructions, constants and alternate-method note

Indian National Physics Olympiad (INPhO)-2026

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION · Tata Institute of Fundamental Research · V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Question Paper

Date: 01 February 2026 · Time: 09:00–12:00 (3 hours) · Maximum Marks: 75

Roll Number: □□□□ – □□□□ – □□□□

Instructions

This booklet consists of 21 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.
Take as many data points as possible for the analysis.
Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.
Except for writing your roll number, no rough work or scribbling is allowed on the question paper.

Table of Constants

Quantity	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
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	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Boltzmann constant	$k_B$	$1.3806\times10^{-23}\ \mathrm{J\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$
One Atmospheric pressure	atm	$1.013\times10^5\ \mathrm{Pa}$

Question Number	1	2	3	4	5	6	Total
Maximum Marks	12	12	11	10	18	12	75

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

Условие

Consider two coaxial conducting circular loops: a lower loop L of radius $a$ and an upper loop U of radius $b$ (see figure). The planes of the loops are parallel and separated by a distance $h$ . The loop U is held fixed in the $x$ - $y$ plane, while the loop L is free to move vertically.

Native coaxial loops U/L, all axes, a,b,h,g, current directions Ia/Ib and both ideal current-source labels.

Each loop is connected to an ideal current source that maintains a constant current $I_a$ in loop L and $I_b$ in loop U. Assume $a\ll b$ , so that the magnetic field produced by the loop U may be treated as uniform over the entire area of the loop L. Let $g$ be the acceleration due to gravity. The system is initially in static equilibrium under gravity.

The loop L is then displaced very slowly by an external agent toward the loop U through a distance $dh$ , while the currents in both loops are maintained constant by ideal power supplies.

During this process, let the change in the gravitational potential energy of the loop L be $dU_g$ , and the change in the energy stored in the magnetic field be $dU_m$ . For the same process, let the extra work done by the power supplies connected to the lower and upper loops be $dW_L$ and $dW_U$ , respectively.

Express each of $dU_g$ , $dU_m$ , $dW_U$ , and $dW_L$ in terms of $I_a$ , $I_b$ , the geometrical parameters $(a,b,h,dh)$ , and any relevant constants, if they are non-zero.

Отговори

Решение

Since the system is initially in static equilibrium under gravity, the upward magnetic force $F_L$ on the lower loop L balances its weight. When the lower loop is displaced slowly upward by a distance $dh$ , the change in its gravitational potential energy is

dU_g=mg\,dh=F_L\,dh.\qquad (2.1)

The lower loop L may be treated as a magnetic dipole with moment

\vec m_L=I_a(\pi a^2)\,\hat z.\qquad (2.2)

The axial magnetic field due to U at a distance $h$ is

B_U(h)=\frac{\mu_0 I_b b^2}{2(b^2+h^2)^{3/2}}.\qquad (2.3)

The force on L due to $B_U$ is

F_{L,z}=\frac{d}{dz}(\vec m_L\cdot\vec B_U),\qquad (2.4)

F_{L,z=-h}=\frac{3\pi}{2}\mu_0 I_a I_b\frac{a^2b^2h}{(b^2+h^2)^{5/2}}.\qquad (2.5)

Substituting Eq. (2.5) into Eq. (2.1) yields

\boxed{dU_g=\frac{3\pi}{2}\mu_0 I_a I_b\frac{a^2b^2h}{(b^2+h^2)^{5/2}}\,dh.}\qquad (2.6)

The magnetic flux through L due to U is

\Phi_L=\pi a^2 B_U(h)=\frac{\mu_0\pi a^2b^2}{2(b^2+h^2)^{3/2}}I_b.\qquad (2.7)

The induced emf in L is

e_L=-\frac{d\Phi_L}{dt},\qquad (2.8)

and the incremental work supplied by L’s power source to maintain current $I_a$ is

|dW_L|=e_L I_a\,dt.\qquad (2.9)

Using Eq. (2.7),

\boxed{dW_L=\frac{3\pi}{2}\mu_0 I_a I_b\frac{a^2b^2h}{(b^2+h^2)^{5/2}}\,dh.}\qquad (2.10)

Equation (2.7) can be written in terms of mutual inductance $M$ as

\Phi_L=MI_b,\qquad M=\frac{\mu_0\pi a^2b^2}{2(b^2+h^2)^{3/2}}.\qquad (2.11)

Using Eq. (2.8),

e_L=-I_b\frac{dM}{dt}.\qquad (2.12)

As the separation changes slowly by $dh$ ,

dM=-\frac{3\mu_0\pi a^2b^2h}{2(b^2+h^2)^{5/2}}\,dh.\qquad (2.13)

The flux through the upper loop due to the lower loop is $\Phi_U=MI_a$ , so an analogous calculation gives

\boxed{|dW_U|=\frac{3\pi}{2}\mu_0 I_a I_b\frac{a^2b^2h}{(b^2+h^2)^{5/2}}\,dh.}\qquad (2.14)

The magnetic energy stored in the two-loop system is

U=\frac12 L_L I_a^2+\frac12 L_U I_b^2+MI_a I_b.\qquad (2.15)

Since the currents are constant, the change in magnetic field energy arises solely from the change in $M$ :

dU_m=I_a I_b\,dM.\qquad (2.16)

Using Eq. (2.13),

\boxed{dU_m=\frac{3\pi}{2}\mu_0 I_a I_b\frac{a^2b^2h}{(b^2+h^2)^{5/2}}\,dh.}\qquad (2.17)

\boxed{dU_g=dW_L=dW_U=dU_m=\frac{3\pi}{2}\mu_0 I_a I_b\frac{a^2b^2h}{(b^2+h^2)^{5/2}}\,dh.}\qquad (2.18)

Source note: the native key prints the minus sign in (2.13), then uses (2.16) to print the positive (2.17) and (2.18). The signed and absolute work notation also varies as printed. These native expressions are preserved rather than silently reconciled.

Теми и трудност

Оценена трудност: 4/5.

Магнетизъм и индукция · Механика · Магнитно поле · Магнитни сили · Електромагнитна индукция · Индуктивност · Работа и енергия

Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

Оригинал в Архива: INPhO2026-Question.pdf · официални решения: INPhO2026-Solution.pdf