Institutt for fysikk, NTNU TFY4155/FY1003 Elektrisitet og magnetisme Vår 2007

Løsningsforslag til øving 12

Veiledning uke 13

Exercise 1

Kirchhoff’s voltage rule (K2) gives

E = \frac{Q}{C} = RI_{R}

while Kirchhoff’s current rule (K1) gives

I = I_{C} + I_{R}

Furthermore, we have

I_{C} = \frac{dQ}{dt}

Hence:

I_{R}(t) = \frac{V_{0}}{R} cos ωt
Q(t) = V_{0}C cos ωt
I_{C}(t) = −ωCV_{0} sin ωt = ωCV_{0} cos(ωt + π/2)

Total current delivered by the source is therefore

I(t) = \frac{V_{0}}{R} cos ωt − ωCV_{0} sin ωt

We want I(t) on the form

I(t) = I_{0} cos(ωt − α)

with amplitude I_{0} = V_{0}/Z, where Z is the impedance of the parallel circuit of R and C, whereas α is the phase constant, i.e., the phase difference between E(t) and I(t). We have

cos(ωt − α) = cos ωt cos α + sin ωt sin α

Hence, by direct comparison:

\frac{cos α}{Z} = \frac{1}{R}
\frac{sin α}{Z} = −ωC

These two equations, with the unknowns Z and α, are easily solved. We find

Z = \frac{R}{\sqrt{1 + (ωRC)^{2}}}

I_{0} = \frac{V_{0}}{R} \sqrt{1 + (ωRC)^{2}}
α = − arctan(ωRC)

In the limit ω → 0 we should find ”well known” results from the DC examples in the lectures, and we do: Z → R and α → 0 so that I_{0} → V_{0}/R. All the current goes through R, whereas the capacitance C now represents an open circuit, transmitting no direct current.

With the given numerical values we have:

ωRC = 2π · 10^{6} · 10 · 16 · 10^{−9} = 1.0

so that

I_{0} = \frac{1.0}{10} · \sqrt{2} = 0.14 A
α = − arctan 1.0 = −45^{∘}

Sketch of E(t), I(t), I_{R}(t) and I_{C}(t):

Sketch of α, I_{0} and Z (with α in radians and ωRC between 0.016 and 160 along the horizontal axis).

Exercise 2

In the first experiment, B = 0. Then, Newton’s second law is:

F = ma = qE
⇒ \frac{dv}{dt} = \frac{q}{m}E
⇒ v(t) = v(0) + \frac{q}{m}Et = \frac{dr}{dt}
⇒ r(t) = r(0) + v(0)t + \frac{q}{2m}Et^{2}

Here, it is natural to choose t = 0 the moment the particle enters the region with E ≠ 0, and furthermore, to choose the origin in this position:

r(0) = (x_{0}, y_{0}) = (0, 0)

Here, the velocity is

v(0) = v x̂

when we orient the x axis towards the right. The y axis is oriented upwards, so that

E = −E ŷ

(i.e., with E > 0) The particle trajectory thus becomes a parabola, just like when we throw a mass in the field of gravity. The velocity in the x direction is not affected by the electric field, so

x(t) = vt

whereas the particle obtains a constant acceleration in the y direction, i.e., the displacement in the y direction, as a function of t, must be determined by

y(t) = −\frac{q}{2m}Et^{2}

The particle will leave the region where E ≠ 0 at the moment

t_{L} = \frac{x(t_{L})}{v} = \frac{L}{v}

The vertical position is then

y(t_{L}) = −\frac{q}{2m}E \frac{L^{2}}{v^{2}}

Already, we may conclude that q < 0 if y(t_{L}) > 0. The distance from x = L to x = L + D is then traveled without influence from any kind of forces, and with a direction relative to the x axis in terms of the angle α, where

tan α = \frac{v_{y}(t_{L})}{v_{x}(t_{L})} = \frac{−\frac{q}{m}E \frac{L}{v}}{v} = −\frac{qEL}{mv^{2}}

Besides, we must have

tan α = \frac{y − y(t_{L})}{D}

where y is where the electron hits the detector, at x = L + D.

The experiment is then repeated with the same E-field, but now we turn on a magnetic field B directed into the plane, so that the particles are no longer deflected by the fields. This implies that the electric force (upwards) is exactly balanced by the magnetic force (downwards). In other words:

F = qE + qv × B = 0
⇒ E = vB
⇒ \frac{1}{v} = \frac{B}{E}

Hence:

\frac{y − y(t_{L})}{D} = −\frac{qEL}{mv^{2}} = −\frac{qEL}{m} · \frac{B^{2}}{E^{2}}
⇒ y + \frac{q}{2m}EL^{2} \frac{B^{2}}{E^{2}} = −\frac{qEL}{m} · \frac{B^{2}}{E^{2}} D
⇒ yE = −\frac{q}{m} · B^{2} (DL + \frac{1}{2}L^{2})
⇒ \frac{q}{m} = −\frac{yE}{B^{2} (DL + \frac{1}{2}L^{2})}

I.e.,

a = \frac{E}{B^{2} (DL + L^{2}/2)}

Exercise 3

a) The speed of the ions when they enter the region with magnetic field is determined by the change in potential energy, going through the voltage difference V, being equal to the change in the kinetic energy of the ions:

eV = \frac{1}{2}mv^{2} ⇒ v = \sqrt{\frac{2eV}{m}}

The centripetal acceleration inside the magnetic field is

a = \frac{v^{2}}{r}

so that Newton’s 2. law gives

F = m\frac{v^{2}}{r} = evB ⇒ r = \frac{mv}{eB}

Radius for the resulting circular path for a particle with mass m becomes

r = \frac{1}{B} \sqrt{\frac{2V m}{e}}

i.e., proportional with \sqrt{m}. Radii and masses for the different isotopes must be related as follows:

\frac{r_{i}}{r_{j}} = \sqrt{\frac{m_{i}}{m_{j}}}

where i, j = 79 or 81. If the points where the ions hit the photographic plate are supposed to be separated by a distance of (at least) a = 1.0 cm, the diameter of the two circular paths must differ by 1.0 cm. We obtain

a = 1.0 cm = 2(r_{81} − r_{79}) = 2r_{79} (\sqrt{\frac{m_{81}}{m_{79}}} − 1)

This gives

r_{79} = \frac{a}{2} (\sqrt{\frac{m_{81}}{m_{79}}} − 1)^{−1} = 0.5 cm · (\sqrt{\frac{81}{79}} − 1)^{−1} ≃ 39.7 cm

and

r_{81} = r_{79} + \frac{a}{2} ≃ 40.2 cm

Now, we can determine how strong magnetic field that can be used to achieve these radii:

B = \frac{1}{r_{81}} \sqrt{\frac{2V m_{81}}{e}} = \frac{1}{0.402} · \sqrt{\frac{2 · 400 · 81 · 1.67 · 10^{−27}}{1.6 · 10^{−19}}} = 0.065 T

This represents the upper limit of B: A stronger magnetic field will reduce both r_{79} and r_{81}, but r_{81} the most, so that the ”hit points” move closer to each other. However, at the same time the diameter d_{81} = 2r_{81} must not be larger than the physical limit of the instrument, given by L = 250cm. That corresponds to a minimum value of the magnetic field strength:

B_{min} = \frac{1}{L/2}\sqrt{\frac{2V m_{81}}{e}} = \frac{1}{1.25} · \sqrt{\frac{2 · 400 · 81 · 1.67 · 10^{−27}}{1.6 · 10^{−19}}} = 0.021 T

In other words, we may use a magnetic field between 21 and 65 mT.