USAPhO 2011, Полуфинал — B1

• a., i. 2000 A
• a., ii. 79.2 MW
• b., i. $B = \frac{\mu_0 I}{2\pi r}$
• b., ii. The loop should be placed vertically, parallel to the wire and directly beneath it, with the long edge vertical.
• b., iii. $\mathcal{E} = \frac{N\mu_0 b}{2\pi}\log\frac{h}{h-a}\, I_0\omega\cos\omega t$
• b., iv. 7760
• c., i. It increases.
• c., ii. 28.1 μF

a.

i. Because the load is purely resistive, the average power is simply

so

ii. The cross-sectional area of the wire is $A = \pi r^2 = 7.07 \times 10^{-4}\ \mathrm{m}^2$, so its resistance is

The power loss is then

b.

i. The field is perpendicular to the wire and to the radius, and from Ampere’s Law

The integral evaluates to $(2\pi r)B$, giving

This well-known result can also be written down without justification.

ii. The induced emf is proportional to the rate of change of the flux through the loop. Since the time dependence of the magnetic field is uniform across space, the rate of change of flux is maximized by maximizing the flux itself. This in turn can be accomplished by maximizing the field in the loop and ensuring that it is normal to the loop. Because the field gets stronger closer to the wire, the loop should be directly below the wire, and since the field is horizontal and perpendicular to the wire at this location, the loop should be vertical and parallel to the wire. Finally, again because the field gets stronger closer to the wire, the long edge of the loop should be vertical.

In summary, the loop should be placed vertically, parallel to the wire and directly beneath it, with the long edge vertical.

iii. Faraday’s law states

where we have dropped the sign, which is not important, and $\Phi_B$ is the magnetic flux through a single loop. The flux is

where we have divided the loop into strips of radial width $dr$ and length $b$. Plugging in the result of part (i),

The time dependence comes only from the current,

Therefore, we have

iv. The RMS value of $I_0\cos\omega t$ is simply $I_{\mathrm{rms}}$, so taking the RMS value of both sides,

where we used $f = \omega/2\pi$. Plugging in the numbers, we have

so the required number of turns is

c.

i. The inductor adds a new component of the current in the wire which is $90^{\circ}$ out of phase with the voltage. The rms current is increased, so the power lost in the wire increases.

This can also be shown more formally using complex impedances. Ohm’s law is

where both $V$ and $I$ are complex numbers, with their relative phase indicating the relative phase of the voltage and current. Without the inductance, $Z = R$. With the inductance, the two impedances add in parallel,

Since $|Z|$ is lower and $V$ remains the same, $|I|$ increases, as argued above.

ii. The most straightforward method is to use complex impedances. We wish to cancel the imaginary component supplied by the inductor, so we need

since they are in parallel. This is equivalent to

so that

One can also arrive at this result without complex impedances. No external current is needed to make an LC circuit oscillate at its natural frequency; current simply sloshes back and forth between the inductor and capacitor, without any going through the power cable. Hence if the power cable frequency matches the natural frequency of the LC circuit, the LC circuit will not affect the current in the cable; the load behaves as if there were no LC circuit present at all. This gives the condition $\omega = 1/\sqrt{LC}$ as above.