CARACTERIZACION DE LA SITUACION AMBIENTAL

Although the prototype target wheel is the same size and shape as the present design for the ILC positron source, the field produced by the magnet cannot reach the levels that may be used in a matching device. Therefore, it is not possible to make measurements of the eddy current effects in the exact conditions that the real target wheel would experience. It is necessary to use simulations to predict the effects: the goal of the experiments using the prototype target wheel is to validate the simulations.

Before discussing the simulations, we develop an analytical model of the system. Although some simplifications are necessary, this will give us an appreciation of the physics, and allow us to judge whether the results from the simulations are reasonable. The problem will be formulated as follows. We consider an infinitely long metal bar, with rectangular cross-section, moving at constant velocityv through a magnetic field. The direction of motion is parallel to the (infinite) length of the bar. The magnetic field is perpendicular to the direction of motion, but the strength varies as a function of position. The situation is illustrated in Fig. 5.7. We choose axes so that the bar is moving in the z direction, and the magnetic field is in the x direction. The edges of the bar are parallel to the x and y axes; let the lengths of the edges in the x and y

directions beax and ay, respectively.

The bar represents the rim of the target wheel. The model neglects effects associated with the fact that the motion is circular, i.e. that different points on the rim are actually moving in different directions, and have some acceleration. However, if the radius of the wheel is large compared to the cross-sectional dimensions of the rim, and compared to the extent of the magnetic field, then these effects should be small. Our goal is to calculate the eddy currents generated in the bar in this model, and then to find the resulting force on the bar. In the case of the target wheel, this force, multiplied by the

y x z ay +V -V V+dV -V-dV I v I I I Bx(z) Bx(z+dz) dz

Figure 5.7: Simplified model of the target wheel moving in a magnetic field, allowing

analytical calculation of the eddy currents and the resulting forces.

radius of the wheel, gives the decelerating torque on the target wheel arising from the eddy currents.

In the simple model shown in Fig.5.7, a voltage is induced in the y direction from the motion of the metal bar through the magnetic field. Consider a short section of the bar of length dz. If the field strength varies as a function of z, then the induced voltage will also vary as a function ofz. Referring to the voltages shown in Fig. 5.7:

2V = v(Bx)_zay, (5.1)

2(V +dV) = v(Bx)z+dzay. (5.2)

Thus, the voltage difference between two points with the samex coordinate, and with

y=ay, but with a (small) separation dz between the zcoordinates, will be:

dV = 1 2v

dBx

dz dz ay. (5.3)

The difference in voltage leads to a current flow in the metal bar, in thezdirection. Since the voltage varies linearly withy, the current densityJ as a function ofyis given by: J(y) =σdV dz y ay/2 =σvdBx dz y, (5.4)

where σ is the conductivity of the material. Then the total current dI in a thin slice of the bar of heightdyand extending across the full width ax of the bar is:

dI(y) =J(y)axdy=σv

dBx

If we imagine closing a current loop with current flowing parallel (or anti-parallel) to they axis, then there will be a force on the section of the bar, resulting from the field acting on the induced current:

d2Fz = (Bx)zdI(y)2y−(Bx)z+dzdI(y)2y=−2

dBx

dz dI(y)y dz. (5.6)

Substituting from Eq. (5.5) gives:

d2Fz =−2σv dBx dz 2 axy2dy dz. (5.7)

The total force on the section of the bar of length dz is found by summing over the force generated by currents at all positions of the bar:

dFz=−2σv dBx dz 2 axdz Z ay/2 0 y2dy=₋ 1 12σv dBx dz 2 axa3ydz. (5.8)

The total force on the bar is obtained by integrating over the entire length of the bar:

Fz=− 1 12σv axa 3 y Z dBx dz 2 dz. (5.9)

We have assumed that the field is uniform in the xand y directions. However, we can generalise the expression for the total force by incorporating the factor ax and one of

the factors ofay into the integral, which then becomes a volume integral:

Fz =− 1 12σv a 2 y Z dBx dz 2 dV. (5.10)

Notice that the force is in the direction opposite to the velocity of the bar: the currents induced by the motion of the bar in the magnetic field act against the motion of the bar. Also notice that the force depends on the square of the derivative of the field along the direction of motion. In a uniform field, there is no force. This is why the fringe field in the target wheel experiment is important. The size of the force also depends on the conductivity of the material (a non-conductor will experience no force, as we would expect) and on the velocity [83].

The power required to keep the bar in motion at constant velocity is readily found from the braking force:

P =₋v Fz = 1 12σv 2_a2 y Z dBx dz 2 dV. (5.11)

This is also equal to the power dissipated in the eddy currents, as may be shown by calculating the ohmic losses directly.

Finally, for the case of a rotating wheel of radiusr, the torque required to keep the wheel rotating at constant angular frequencyω =v/r is given by:

τ =₋r Fz = 1 12σωr 2_a2 y Z dBx dz 2 dV. (5.12)

Notice that the torque has a strong dependence on the size of the rim: assuming that the rim is fully immersed in the field, and that the field is approximately uniform in the radial direction, the torque varies as the cube of the width of the rim (i.e.

τ _∝a3_y). In the case of the target wheel, the rim is connected to the drive shaft by five spokes. At each spoke, the value ofay is large compared to the value where there is no

spoke. Therefore, although each spoke is quite thin, the spokes may make a significant contribution to the total (average) torque. If we estimate the torque using Eq. (5.12) neglecting the spokes, then we expect that we will underestimate the true value.

Nevertheless, let us apply the above results to our positron target prototype, to give some feeling for the values we might expect for the torque and the power dis- sipation from the eddy currents. The material Ti-6%Al-4%V has a conductivity of 5.8_×106_Ω−1_m−1_{. The target rim width a}

y is approximately 30 mm, and target rim

thicknessaxis equal to 15.6 mm. Let us consider the “worst” case: revolution frequency

of 2000 rpm (rim velocity approximately 100 m/s); peak field of 1.45 T, and immersion depth of 50.25 mm. In this case, performing the integral under the derivative of the curve shown in Fig. 5.6gives:

Z

dBx

dz

2

dV _≈0.019 T2m. (5.13)

Then we find that the torque is approximately 43 Nm, and the power dissipated is approximately 9 kW.