A large solenoid coil for a physics experiment is made of a...

A large solenoid coil for a physics experiment is made of a single layer of conductor of cross section $4 \mathrm{~cm} \times 2 \mathrm{~cm}$ with a cooling water hole 2 cm $\times 1 \mathrm{~cm}$ in the conductor. The coil, which consists of 100 turns, has a diameter of 3 meters, and a length of 4 meters (the insulation thickness is negligible). At the two ends of the coil are circular steel plates to make the field uniform and to return the magnetic flux through a steel cylindrical structure external to the coil, as shown in the diagram. A magnetic field of 0.25 Tesla is desired. The conductor is made of aluminium. (a) What power (in kilowatts) must be supplied to provide the desired field, and what must be the voltage of the power supply? (b) What rate of water flow (litres/second) must be supplied to keep the temperature rise of the water at $40^{\circ} \mathrm{C}$ ? Neglect all heat losses from the coil except through the water. (c) What is the outward pressure exerted on the coil by the magnetic forces? (d) If the coil is energized by connecting it to the design voltage calculated in (a), how much time is required to go from zero current to $99 \%$ of the design current? Neglect power supply inductance and resistance. The resistivity of aluminium is $3 \times 10^{-8}$ ohm-meters. Assume that the steel is far below saturation. FIGURE CAN'T COPY.

A large solenoid coil for a physics experiment is made of a single layer of conductor of cross section $4 \mathrm{~cm} \times 2 \mathrm{~cm}$ with a cooling water hole 2 cm $\times 1 \mathrm{~cm}$ in the conductor. The coil, which consists of 100 turns, has a diameter of 3 meters, and a length of 4 meters (the insulation thickness is negligible). At the two ends of the coil are circular steel plates to make the field uniform and to return the magnetic flux through a steel cylindrical structure external to the coil, as shown in the diagram. A magnetic field of 0.25 Tesla is desired. The conductor is made of aluminium.
(a) What power (in kilowatts) must be supplied to provide the desired field, and what must be the voltage of the power supply?
(b) What rate of water flow (litres/second) must be supplied to keep the temperature rise of the water at $40^{\circ} \mathrm{C}$ ? Neglect all heat losses from the coil except through the water.
(c) What is the outward pressure exerted on the coil by the magnetic forces?
(d) If the coil is energized by connecting it to the design voltage calculated in (a), how much time is required to go from zero current to $99 \%$ of the design current? Neglect power supply inductance and resistance. The resistivity of aluminium is $3 \times 10^{-8}$ ohm-meters. Assume that the steel is far below saturation.
FIGURE CAN'T COPY.

Key Concepts

Magnetic Field of a Solenoid

This concept involves understanding that a solenoid—a long, coiled conductor—produces a near?uniform magnetic field inside its volume when an electric current flows through it. The magnetic field strength is typically calculated using the relationship B = ??nI, where ?? is the permeability of free space, n is the number of turns per unit length, and I is the current. This is foundational for designing electromagnets that achieve a specific field strength.

Electrical Resistance and Joule Heating

Electrical resistance quantifies how the conductor opposes the flow of electric current. When current flows through the resistance of the coil, electrical energy is dissipated as heat following the Joule heating law (P = I²R). Determining the resistance of the conductor and the associated power loss is essential for specifying the required voltage and ensuring that the heating can be managed safely.

Thermal Management and Cooling Flow Rate Calculation

This concept pertains to the removal of heat generated by Joule heating in the coil. By calculating the heat energy that must be removed to keep the conductor within safe operating temperatures, and knowing the specific heat capacity of the coolant (water, in this case), one can determine the necessary flow rate. Effective thermal management is crucial to prevent overheating in high-power electrical devices.

Magnetic Pressure in Electromagnetic Devices

Magnetic pressure arises from the energy density of the magnetic field and represents the mechanical force exerted by the magnetic field on the conductors of the coil. It is often calculated using the expression P = B²/(2??), which relates the magnetic field strength to the pressure. This concept is important in assessing structural integrity and ensuring the mechanical design can withstand these forces.

Transient Response of RL Circuits

When an electric circuit includes an inductor along with resistive elements, its current does not instantaneously reach its steady?state value when a voltage is applied. Instead, it increases following an exponential behavior characterized by a time constant (? = L/R). This concept is key to understanding how quickly a coil can reach a desired current and thus achieve the intended magnetic field strength.

Transcript

00:01 Our question wants us to consider an electro -magnetic coil with a diameter that is made from square wire.

00:09 The square wire has a length of 2 millimeters.

00:13 So l here is 0 .02 meters.

00:16 And the diameter of the coil is 2 meters.

00:21 So d here is 2 .0 meters.

00:24 Okay.

00:26 We're also told that the power supply has a voltage v of 35 volts and a maximum power output of 1 .0.

00:31 Kilowatts, which i call p of 1 .0 times 10 of the third watts.

00:36 We know the copper wire here, or the wire here is a copper wire.

00:43 So i also write the resistivity of the copper as a row.

00:46 1 .68 times 10 to the minus 8 oms per meter.

00:51 For part a, we're asked how many turns are needed to run the power supply at maximum power? well, the first thing we're to consider is the equation for power.

01:01 So power here, and let's indicate this is a second.

01:03 Part a first power here is equal to the voltage squared v divided by the resistance r where r the resistance is equal to the resistivity row times the total length here not just the the length l which was the the length of the square copper wire we're talking about the total length l which is the length of the wire when the voltage power is maximum here divided by the area a therefore we can express the length of the wire given these two equations from the voltage at maximum power as l is equal to we'll use capital l here distinguish it from little l l is equal to the voltage squared times the area divided by the power p times the resistivity row here let's make sure we have enough room actually, there we go, power p times resistivity row.

02:20 Well, in the number of turns here is equal to l, the total length divided by two times pi times the radius, but the radius is d over two, right? since we were given the diameter.

02:42 So we just have the expression for l on the left side there so we can replace that and simplify a little bit.

02:50 So this is v squared times the area divided by pi because we can cancel that too with the one -half, pi times, pi times d, actually, not times r.

03:10 There we go.

03:12 In the area here, since it's a rectangular or a square, is just the square of little l...

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