USAPhO 2005, Полуфинал — B1

a. (7 т.)

By symmetry, the $x$ and $y$ components of the electric field from different parts of the ring will cancel:

Therefore, we need to add up the $z$-components of the electric field from the various parts of the ring:

The electric field due to an infinitesimal charge $dq$ on the ring at a distance $z$ from the origin is

where $r=(b^2+z^2)^{1/2}$. (B1-2)

($r$ is the distance from a point on the ring to the point $(0,0,z)$.) $\hat r$ is a unit vector at the point $(0,0,z)$ pointing away from the charge $dq$ on the ring.

The $z$-component is then given by:

where $\theta$ is the angle between $\hat r$ and the $z$ axis.

Substituting eq. (B1-1) into (B1-2) yields:

Integrating (B1-4) around the ring and observing that $r$ is the same for all points on the ring yields

Using the fact that $E_x=E_y=0$ and substituting (B1-2) into (B1-5) gives us the answer.

b. (5 т.)

The potential a distance $r$ away from an infinitesimal charge $dq$ is given by

Integrating around the ring, using the fact that $r$ is the same from every point on the ring to the point $(0,0,z)$, and substituting (B1-2) for $r$ yields

Note that it is easier to solve part (b) first and then to solve part (a) using

c. (10 т.)

The potential energy of the system is given by

Using (B1-8),

Factoring out $b^2$,

Using the fact that $\lvert z\rvert\ll b$ and using the approximation $(1+x)^n\approx1+nx$ when $x\ll1$, (B1-12) becomes

d. (5 т.)

Mechanical energy is conserved, so

Since the charge is at rest at $z=z_0$ and using (B1-13),

Solving for $v(0)$,

e. (8 т.)

We shall use Newton’s Second Law:

The force on the charge $-q$ is $-qE$ where $E$ is given by (B1-6). Substituting into (B1-17),

But, $z\ll b$, so $(b^2+z^2)^{3/2}\approx b^3$ and, after dividing both sides by $m$, (B1-18) becomes

We recognize that the differential equation

is simple harmonic motion. The solution to (B1-20) is

Since $z(0)=z_0$, we find that $A=z_0$.

By comparison with (B1-19), we find that

We can derive an expression for the velocity as a function of time by differentiating (B1-21):

(where $\omega$ is given by (B1-22).)

Note that an alternative method of solving part (d) is to note that the speed at the origin will be the maximum value of the magnitude of the velocity

Note that an alternative (easier) method of solving part (e) is to note that the potential energy found in (B1-13) is of the form $U(z)=U(0)+kz^2/2$.

Therefore, since it starts from rest, we know that $z(t)=A\cos\omega t$ and that

Thus, $\omega=v_{\max}/z_0$ and $v_{\max}$ was found in (B1-16).

f. (10 т.)

According to the Biot-Savart law, the magnetic field due to a moving infinitesimal charge $dq$ is given by

Since $d\vec B$ depends on the cross product $\vec v\times\hat r$, $d\vec B$ must be perpendicular to both $\vec v$ and $\hat r$.

By symmetry, when we add the magnetic field vectors due to all points on the rotating ring, the $x$ and $y$ components will cancel. Therefore, we need to find the $z$ components and add them. Let $\phi$ be the angle between $\vec B$ and $\hat z$.

Substituting $v=b\omega$ and (B1-24) into (B1-25):

Integrating (B1-26) around the ring and observing that $r$ is the same for all points on the ring yields

Using the fact that $B_x=B_y=0$ and substituting (B1-2) into (B1-27) gives us the answer.

g. (5 т.)

A point charge $-q$ moving with velocity $\vec v$ in a magnetic field $\vec B$ experiences a magnetic force

However, since the charge is moving in the $-\hat k$ direction and the magnetic field is in the $\hat k$ direction, $\vec v\times\vec B=0$, so $\vec F_{\rm magn}=0$.