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Условие Отговори Решение

IZhO 2020, Теория — Задача 3. Ring in a magnetic field

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XVI International Zhautykov Olympiad/Theoretical Competition · 10 януари 2020 г. · 10 т.

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THEORETICAL COMPETITION

January 10, 2020

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The time available for the theoretical competition is 4 hours. There are three questions.
Use only the pen provided.
You can use your own calculator for numerical calculations. If you don’t have one, please ask for it from Olympiad organizers.
You are provided with Writing sheet and additional paper. You can use the additional paper for drafts of your solutions but these papers will not be checked. Your final solutions which will be evaluated should be on the Writing sheets. Please use as little text as possible. You should mostly use equations, numbers, figures and plots.
Use only the front side of Writing sheets. Write only inside the boxed area.
Begin each question on a separate sheet of paper.
Fill the boxes at the top of each sheet of paper with your country (Country), your student code (Student Code), the question number (Question Number), the progressive number of each sheet (Page Number), and the total number of Writing sheets used (Total Number of Pages). If you use some blank Writing sheets for notes that you do not wish to be evaluated, put a large X across the entire sheet and do not include it in your numbering.
At the end of the exam, arrange all sheets for each problem in the following order:
- Used Writing sheets in order.
- The sheets you do not wish to be evaluated.
- Unused sheets.
- The printed problems.
Place the papers inside the envelope and leave everything on your desk. You are not allowed to take any paper out of the room.

SOLUTIONS TO THE PROBLEMS OF THE THEORETICAL COMPETITION

SOLUTIONS TO THE PROBLEMS OF THE THEORETICAL COMPETITION

Attention. Points in grading are not divided!

Условие

Uniformly charged ring

A very thin ring of mass $m$ and radius $r$ is uniformly charged along its length with a charge $q$ . At the initial moment of time, the ring rests horizontally and is released without a push. The subsequent motion of the ring appears in the vertical gravitational field of the Earth, characterized by the acceleration of gravity $g$ and in the horizontal radial magnetic field of induction $B$ . Neglect air resistance and assume that the plane of the ring remains horizontal at all times.

A horizontal thin ring with radial arrows of the magnetic field B pointing outwards and a vertical arrow g pointing downwards; labels B and g.

3.1 Find the maximum velocity of the ring center of mass $v_{max}$ for the entire time of motion. [3,2 т.]

3.2 Find the time interval $\Delta t$ elapsed from the start of the ring motion to its first reaching of the maximum velocity of the center of mass. [0,8 т.]

3.3 Find the maximum height $h_{max}$ at which the ring center of mass falls over the entire time of motion. [0,2 т.]

Conductive ring

A very thin ring of mass $m$ and radius $r$ is made of a conductive material with a resistivity $\rho$ and a cross section area $s \ll r^2$ . At the initial time moment $t = 0$ the ring rests horizontally and is released without a push. The subsequent motion of the ring appears in the vertical gravitational field of the Earth, characterized by the acceleration of gravity $g$ and in the horizontal radial magnetic field of induction $B$ . Neglect air resistance and assume that the plane of the ring remains horizontal at all times.

3.4 Find the steady-state velocity $v_0$ of the ring center of mass after a sufficiently large period of time having passed. [1,8 т.]

3.5 The dependence of the current strength $I(t)$ in the ring on time t has the following form

I(t) = A_1 + B_1\exp(\gamma_1 t).

Find the constants $A_1, B_1$ and $\gamma_1$ . [1 т.]

Conductive ring with a cut

A very thin ring of mass $m$ and radius $r$ is made of a conductive material with a resistivity $\rho$ and a cross section area $s$ . A cut with a width $\delta \ll \sqrt{s} \ll r$ was made along the radius of the ring. At the initial time moment $t = 0$ the ring rests horizontally and is released without a push. The subsequent motion of the ring appears in the vertical gravitational field of the Earth, characterized by the acceleration of gravity $g$ and in the horizontal radial magnetic field of induction $B$ . Neglect air resistance and assume that the plane of the ring remains horizontal at all times.

A horizontal thin ring with a narrow radial cut of width δ, radial arrows of the magnetic field B and a vertical arrow g pointing downwards; labels B, δ and g.

3.6 Find the steady-state acceleration $a_0$ of the ring center of mass after a sufficiently large period of time having passed. [1 т.]

3.7 The dependence of the current strength $I(t)$ in the ring on time t has the following form

I(t) = A_2 + B_2\exp(\gamma_2 t).

Find the constants $A_2, B_2$ and $\gamma_2$ . [2 т.]

Отговори

Решение

Problem 3. Ring in a magnetic field (10.0 points)

Uniformly charged ring

3.1 Under the action of gravity, the center of mass of the ring acquires a velocity $v$ directed vertically downward. In this case, the Lorentz force $F_{L1}$ arises, leading to the rotation of the ring around its own axis with an angular velocity $\omega$ , which in turn leads to the appearance of the vertical component of the Lorentz force $F_{L2}$ directed against gravity, regardless of the sign of the ring charge.

The equation of motion of the ring center of mass has the form

m\frac{dv}{dt} = mg - F_{L2}, \qquad (1)

where the Lorentz force is written as

F_{L1} = qv_{rot}B, \qquad (2)

and the linear speed of ring rotation

v_{rot} = \omega r. \qquad (3)

Thus, the equation of motion of the ring center of mass is finally derived as

m\frac{dv}{dt} = mg - q\omega rB. \qquad (4)

The equation of rotational motion of the ring is written as

I\frac{d\omega}{dt} = M_{L1}, \qquad (5)

where the torque of the Lorentz force moment $F_{L1}$ is determined by the expression

M_{L1} = qvBr, \qquad (6)

and the moment of inertia of the ring is equal to

I = mr^2. \qquad (7)

Let $h$ be the vertical displacement of the ring center of mass, then its speed is

v = \frac{dh}{dt}. \qquad (8)

Putting together (5), (6) and (8) and integrating over time, taking into account the initial condition $\omega = 0$ at $h = 0$ , we obtain the relation

I\omega = qBrh. \qquad (9)

At the moment when the speed of the ring center of mass is maximum, the total force on the right side of equation (4) vanishes, which leads to the expression

mg = qBr\omega_0. \qquad (10)

Using relation (9) for this particular moment in time

I\omega_0 = qBrh_0, \qquad (11)

we apply the law of energy conservation in the following form

mgh_0 = \frac{mv_{max}^2}{2} + \frac{I\omega_0^2}{2}. \qquad (12)

Solving equations (10)-(12) together with the expression for the moment of inertia (7), we find the maximum velocity of the ring center of mass in the following form

v_{max} = \frac{mg}{qB}. \qquad (13)

3.2 Substituting relation (9) into the equation (4) of motion of the ring center of mass and using (8), we obtain the equation of harmonic oscillations

m\frac{d^2h}{dt^2} = mg - \frac{(qB)^2}{m}h \qquad (14)

with the frequency

\omega_L = \frac{qB}{m}. \qquad (15)

The time sought is quarter of the period of oscillation, i.e.

\Delta t = \frac{\pi}{2\omega_L} = \frac{\pi m}{2qB}. \qquad (16)

3.3 The initial velocity of the ring center of mass is zero and reaches its maximum at the moment of passage of the equilibrium position, therefore, the maximum height $h_{max}$ by which the ring center of mass descends is obtained as

h_{max} = 2h_0 = \frac{2gm^2}{q^2B^2}. \qquad (17)

Conductive ring

3.4 Under the action of gravity, the ring center of mass acquires a velocity $v$ directed vertically downward. In this case, an induction current $I$ arises in the ring as a result of the action of a magnetic field, which leads to the appearance of a vertical Lorentz force $F_L$ directed against gravity.

The equation of motion of the ring center of mass has the form

m\frac{dv}{dt} = mg - F_L, \qquad (18)

and the Lorentz force is determined by the expression

F_L = BIL, \qquad (19)

with the ring length

L = 2\pi r. \qquad (20)

When moving in a magnetic field, the following electromotive force arises in the ring

\mathcal{E} = \frac{d\Phi}{dt} = BLv, \qquad (21)

which, according to Ohm's law, leads to the appearance of the induction current of strength

\mathcal{E} = IR, \qquad (22)

where the ring resistance is

R = \rho\frac{L}{s}. \qquad (23)

In the steady-state fall mode of the ring center, its velocity $v = v_0$ remains unchanged, then from (18)-(23) we obtain

v_0 = \frac{mg\rho}{2\pi r s B^2}. \qquad (24)

3.5 Expressing the velocity from (21), (22) and substituting it into equation (18), as well as using (19), we obtain the differential equation

\frac{mR}{BL}\frac{dI}{dt} = mg - BLI, \qquad (25)

with the initial condition

I(0) = 0. \qquad (26)

The solution to equation (25) when (26) is satisfied is the function

I(t) = \frac{mg}{2\pi rB}\left[1 - \exp\left(-\frac{2\pi r s B^2}{m\rho}t\right)\right]. \qquad (27)

Whence,

A_1 = \frac{mg}{2\pi rB}, \qquad (28)

B_1 = -\frac{mg}{2\pi rB}, \qquad (29)

\gamma_1 = -\frac{2\pi r s B^2}{m\rho}. \qquad (30)

Conductive ring with a cut

3.6 The equation of motion of the ring center of mass of the is still described by equations (18)-(20), and an electromotive force (21) is also generated in the ring. However, in this case, charges of the opposite sign accumulate at the cut edges; therefore, instead of Ohm's law (22), we have

\mathcal{E} - \frac{q}{C} = IR, \qquad (31)

where

C = \frac{\varepsilon_0 S}{\delta}. \qquad (32)

Since the cut edges are charged by induction current, then

I = \frac{dq}{dt}. \qquad (33)

In the steady state, the acceleration of the ring center of mass is constant, so according to the equation of motion (18) and (19), the current strength is also constant. Differentiating (31) with (33) and (21) taken into account, we finally obtain the steady-state acceleration

a_0 = \frac{g}{\left(1 + \frac{B^2(2\pi r)^2\varepsilon_0 s}{m\delta}\right)}. \qquad (34)

3.7 Differentiating (31) with (21) and (33) taken into account, we obtain

BL\frac{dv}{dt} = \frac{I}{C} + R\frac{dI}{dt}. \qquad (35)

Dividing this equation by the equation of motion (18) and substituting (19), we obtain the differential equation for the current in the ring

R\frac{dI}{dt} = gBL - \left(\frac{1}{C} + \frac{B^2L^2}{m}\right)I \qquad (36)

with the initial condition

I(0) = 0. \qquad (37)

The solution of equation (36) with (37) is the function

I(t) = \frac{2\pi r g\varepsilon_0 s B}{\delta\left(1 + \frac{B^2(2\pi r)^2\varepsilon_0 s}{m\delta}\right)}\left[1 - \exp\left(-\left(1 + \frac{B^2(2\pi r)^2\varepsilon_0 s}{m\delta}\right)\frac{\delta}{2\pi r\rho\varepsilon_0}t\right)\right]. \qquad (38)

Hence,

A_2 = \frac{2\pi r g\varepsilon_0 s B}{\delta\left(1 + \frac{B^2(2\pi r)^2\varepsilon_0 s}{m\delta}\right)}, \qquad (39)

B_2 = -\frac{2\pi r g\varepsilon_0 s B}{\delta\left(1 + \frac{B^2(2\pi r)^2\varepsilon_0 s}{m\delta}\right)}, \qquad (40)

\gamma_2 = -\left(1 + \frac{B^2(2\pi r)^2\varepsilon_0 s}{m\delta}\right)\frac{\delta}{2\pi r\rho\varepsilon_0}. \qquad (41)

	Content	Points
3.1	Equation (1): $m\frac{dv}{dt} = mg - F_{L2}$	0,3	3,2
	Formula(2): $F_{L1} = qv_{rot}B$	0,2
	Formula (3): $v_{rot} = \omega r$	0,2
	Equation (4): $m\frac{dv}{dt} = mg - q\omega rB$	0,2
	Equation (5): $I\frac{d\omega}{dt} = M_{L1}$	0,3
	Formula(6): $M_{L1} = qvBr$	0,2
	Formula (7): $I = mr^2$	0,2
	Formula (8): $v = \frac{dh}{dt}$	0,2
	Formula (9): $I\omega = qBrh$	0,4
	Formula (10): $mg = qBr\omega_0$	0,4
	Formula (12): $mgh_0 = \frac{mv_{max}^2}{2} + \frac{I\omega_0^2}{2}$	0,3
	Formula (13): $v_{max} = \frac{mg}{qB}$	0,3
3.2	Equation (14): $m\frac{d^2h}{dt^2} = mg - \frac{(qB)^2}{m}h$	0,3	0,8
	Formula (15): $\omega_L = \frac{qB}{m}$	0,2
	Formula(16): $\Delta t = \frac{\pi}{2\omega_L} = \frac{\pi m}{2qB}$	0,3
3.3	Formula(17): $h_{max} = 2h_0 = \frac{2gm^2}{q^2B^2}$	0,2	0,2
3.4	Equation (18): $m\frac{dv}{dt} = mg - F_L$	0,3	1,8
	Formula (19): $F_L = BIL$	0,2
	Formula (20): $L = 2\pi r$	0,2
	Formula (21): $\mathcal{E} = \frac{d\Phi}{dt} = BLv$	0,3
	Formula (22): $\mathcal{E} = IR$	0,3
	Formula (23): $R = \rho\frac{L}{s}$	0,2
	Formula (24): $v_0 = \frac{mg\rho}{2\pi r s B^2}$	0,3
3.5	Equation (25): $\frac{mR}{BL}\frac{dI}{dt} = mg - BLI$	0,2	1,0
	Condition (26): $I(0) = 0$	0,2
	Formula (28): $A_1 = \frac{mg}{2\pi rB}$	0,2
	Formula (29): $B_1 = -\frac{mg}{2\pi rB}$	0,2
	Formula (30): $\gamma_1 = -\frac{2\pi r s B^2}{m\rho}$	0,2
3.6	Equation (31): $\mathcal{E} - \frac{q}{C} = IR$	0,3	1,0
	Formula (32): $C = \frac{\varepsilon_0 S}{\delta}$	0,2
	Formula (33): $I = \frac{dq}{dt}$	0,2
	Formula (34): $a_0 = \frac{g}{\left(1 + \frac{B^2(2\pi r)^2\varepsilon_0 s}{m\delta}\right)}$	0,3
3.7	Equation (35): $BL\frac{dv}{dt} = \frac{I}{C} + R\frac{dI}{dt}$	0,5	2,0
	Equation (36): $R\frac{dI}{dt} = gBL - \left(\frac{1}{C} + \frac{B^2L^2}{m}\right)I$	0,5
	Condition (37): $I(0) = 0$	0,1
	Formula (39): $A_2 = \frac{2\pi r g\varepsilon_0 s B}{\delta\left(1 + \frac{B^2(2\pi r)^2\varepsilon_0 s}{m\delta}\right)}$	0,3
	Formula (40): $B_2 = -\frac{2\pi r g\varepsilon_0 s B}{\delta\left(1 + \frac{B^2(2\pi r)^2\varepsilon_0 s}{m\delta}\right)}$	0,3
	Formula (41): $\gamma_2 = -\left(1 + \frac{B^2(2\pi r)^2\varepsilon_0 s}{m\delta}\right)\frac{\delta}{2\pi r\rho\varepsilon_0}$	0,3
Total			10,0

Mathematical hints for the theoretical problems

Mathematical hints for the theoretical problems

The following integrals may be useful:

\int\frac{dx}{ax+b} = \frac{1}{a}\ln|ax+b|.

\int x^n dx = \frac{x^{n+1}}{n+1},\ \text{where } n \text{ is integer}

(1+x)^{\gamma} \approx 1 + \gamma x + \frac{\gamma(\gamma-1)}{2}x^2,\ \textit{for } x \ll 1 \textit{ and any } \gamma

Оригинал в Архива: IZhO-2020-Theory_eng.pdf · официални решения: IZhO-2020-Theory_eng_sol.pdf