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IZhO 2006, Теория — Задача 3

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Всички материали от IZhO 2006

2nd INTERNATIONAL ZHAUTYKOV PHYSICS AND MATHEMATICS OLYMPIAD. KAZAKHSTAN, ALMATY 2006. Physics, Theoretical competition. · 12 т.

Условие

Given a ring of radius $R$ with a current $I$ flowing in it.

A current ring of radius R (drawn as an ellipse) with current I shown by an arrow; dashed lines from the top and bottom of the ring converge at the point O1 on the dash-dotted axis, the angle between them marked 2α. — Fig.3.1

1. Calculate the magnetic field at point $O_1$ located on the axis of the ring. The total angular span of the ring at point $O_1$ equals $2\alpha$ . (See Fig.3.1)

The solenoid of radius $R$ and length $l$ consists of $N$ coils, distributed uniformly. The current through the solenoid equals $I$ .

2. Find the magnetic field on the axis of the solenoid at the point, from which the angular span of the ends of solenoid equals $2\alpha$ and $2\beta$ . (See Fig.3.2)

A solenoid (rows of small coils top and bottom) with a dash-dotted axis; from the point O on the axis outside the solenoid, dashed lines to the near end form the angle 2α and dashed lines to the far end form the angle 2β. — Fig.3.2

A solenoid with field B0 inside on the dash-dotted axis; a point outside on the axis at distance x from the end (marked with a brace) where the field is 0,1B0. — Fig.3.3

In the following we assume that $\ell \gg R$ .

3. a) Calculate the magnetic field $B_0$ inside the solenoid far away from its ends.

3. в) Find the distance $x$ , for which $B = 0.1\cdot B_0$ (See Fig.3.3)

3. c) Calculate the inductance of the coil $L$ , assuming that magnetic field inside the solenoid far away from its ends is uniform.

Magnetic bullet is moving along the axis of the solenoid. Solenoid is connected to a capacitor $C$ . Magnetic moment of the bullet $M$ is parallel to the axis of the solenoid. We will neglect the change of the velocity of the bullet.

4. a) Write the condition, that the time it takes the bullet to pass the region of non uniform magnetic field is much smaller than the period of oscillations in LC circuit. Assume in the following, that this condition is always satisfied.

4. b) Find the values of the bullet velocities $\upsilon$ , such that the amplitude of oscillations in LC circuit after the bullet passes the solenoid is maximal.

4. c) What is the value of the current oscillation amplitude $I_{\text{мах}}$ for this case? Draw the graph of $I(t)$ .

4. d) Prove that the magnetic force acting on a bullet equals $M\dfrac{\partial B}{\partial x}$ and is parallel to the axis of the solenoid.

Comment:

You may consider the bullet to be a small ring of area $S_0$ ,with a current $I_0$ , and $M = S_0I_0$ .

In the theory of the magnetism the following theorem is proven: If one denotes the flux of the magnetic field created by the first contour through the second contour as $L_{12}I_1$ , and the flux of the magnetic field created by the second contour through the first contour as $L_{21}I_2$ ,then $L_{12} = L_{21}$ . It is assumed that the positive direction of the flux is consistent with the positive direction of the current in each contour .

Отговори

Решение

a) From the symmetry consideration the resulting magnetic field on the axis is directed along the axis. According to the Bio-Savaro, the field of a small element of a current $I$ with a length $R\cdot d\varphi$ is directed at an angle $\pi/2 - \alpha$ to the axis and is equal to

\frac{\mu_0 I d\varphi}{4\pi R}\sin^2\alpha \qquad (2.1)

By projecting it on the axis and integrating over the $\varphi$ from 0 to 2π we get

B = \frac{\mu_0 I}{2R}\sin^3\alpha \qquad (2.2)

(1.0 point)

Geometry for the solenoid field: a strip of coils of width dx at height R above the dash-dotted axis; dashed lines from the strip to a point on the axis make the angle θ with the axis and enclose the small angle dθ; the horizontal distance x is marked with a brace. — Fig.3.4

b) Since $x = Rctg\theta$ (see fig..3.4), the length of a strip defined by angles θ and θ+dθ is

dx = \frac{R}{\sin^2\theta}d\theta. \qquad (2.3)

Using this and the result of the previous item we find

B = \int\frac{\mu_0 NI}{2lR}\sin^3\theta dx = \frac{\mu_0 NI}{2l}\int\limits_{\alpha}^{\beta}\sin\theta d\theta = \frac{\mu_0 NI}{2l}(\cos\beta - \cos\alpha). \qquad (2.4)

(2 points)

c) By formula (2.4) with $\cos\beta \approx 1$ и $\cos\alpha \approx -1$ we have

B_0 = \frac{\mu_0 NI}{l} \qquad (2.5)

(0,5 points)

d) Inserting $\cos\beta \approx 1$ into (2.4) we get an equation for $\cos\alpha$

1 - \cos\alpha = 0{,}2, \qquad (2.6)

whence

x = 4R/3 \qquad (2.7)

(1.0 point)

e) The total flux through the solenoid is equal to

\Phi = B_0SN = \mu_0\frac{\pi R^2N^2}{l}I, \qquad (2.9)

hence

L = \mu_0\frac{\pi R^2N^2}{l} \qquad (2.10)

(0,5 points)

\frac{\upsilon\cdot\sqrt{LC}}{R} \gg 1 \qquad (2.11)

(1.0 point)

g) Consider the variation of the flux of the magnetic field through the bullet during its flight. Using the theorem of the comment we have

\Phi = L_{12}I_0 = \frac{B(x,I)S_0}{I}I_0 = \frac{B(x,I)}{I}M \qquad (2.12)

(1.0 point)

where $B(x,I)$ is the induction of the field of the solenoid, I – its current. Since $B(x,I)/I$ doesn’t depend on I, the flux through the bullet changes only when it flies through the nonuniform region of the field. According to (2.11) the total flux through the solenoid doesn’t change during that time of flight, and hence from the flux conservation we find that the current jump $I_1$ induced in the solenoid would be defined by equation

LI_1 = M\frac{\mu_0 N}{l}. \qquad (2.13)

(1.0 point)

Inserting L from (2.10) we get

I_1 = \frac{M}{\pi R^2N}. \qquad (2.14)

After the bullet’s emergence from the nonuniform region the magnetic field flux through it doesn’t vary, hence in the current in the solenoid oscillates harmonically. When the bullet arrives to the exit from the solenoid, it again enters a nonuniform region of the magnetic field and another sudden change of the current, with a sign opposite to that at the entrance into the solenoid, occurs. To have a maximal residual current in the oscillator it should have the same sign as the current in the circuit, hence the number of oscillations of the contour must be equal odd number of half oscillations. Hence

\upsilon = \frac{l}{(2n-1)\pi\sqrt{LC}},

где n=1, 2, 3,… (2.15) (1.0 point)

Graph of I/I1 versus t/T from 0 to 2,5 (vertical axis from -2,5 to 2,5): the current starts at 1, falls as a cosine to -1 at t/T = 0,5, jumps to -2, then oscillates with amplitude 2 (maximum 2 at t/T = 1, minimum -2 at 1,5, back to 2 at 2). — Fig.3.5

I_{\max} = 2I_1 = \frac{2M}{\pi NR^2} \qquad (2.16)

(0,5 points)

(1.0 point) (graph, Fig.3.5)

i) The Ampere force acting on the bullet is $2\pi B_{\perp}I_0R$ . Using the fact that the flux of the magnetic field through the closed surface is zero we find that

-(B_x - B_{x+dx})\pi R^2 = B_{\perp}2\pi Rdx,

(0,5 points)

B_{\perp}2\pi RI_0 = -\frac{B_x - B_{x+dx}}{dx}\pi R^2I_0,

(0,5 points)

F = M\frac{\partial B}{\partial x}

(0,5 points)

Оригинал в Архива: IZhO-2006_╨óheory_eng.doc · официални решения: IZhO-2006_╨óheory_eng_sol.doc