Problem 90 - University Physics with Modern Physics

Question

The Electromagnetic Pump. Magnetic forces acting on conducting fluids provide a convenient means of pumping these fluids. For example, this method can be used to pump blood without the damage to the cells that can be caused by a mechanical pump. A horizontal tube with rectangular cross section (height \(h,\) width w) is placed at right angles to a uniform magnetic field with magnitude \(B\) so that a length \(l\) is in the field (Fig. 27.75). The tube is filled with a conducting liquid, and an electric current of density \(J\) is maintained in the third mutually perpendicular direction. (a) Show that the difference of pressure between a point in the liquid on a vertical plane through \(a b\) and a point in the liquid on another vertical plane through \(c d,\) under conditions in which the liquid is prevented from flowing, is \(\Delta p=J I B\) . (b) What current density is needed to provide a pressure difference of 1.00 atm between these two points if \(B=2.20 \mathrm{T}\) and \(l=35.0 \mathrm{mm} ?\)

Step-by-Step Solution

Answer

(a) \(\Delta p = J l B\). (b) \( J \approx 1314796 \text{ A/m}^2 \).

Step 1: Understanding Lorentz Force

The Lorentz force, which acts on charges moving in a magnetic field, is given by \( F = q(\mathbf{v} \times \mathbf{B}) + q\mathbf{E} \). For a current flowing perpendicular to a magnetic field, the force per unit length on the fluid is \( f = J l B \), where \( J \) is the current density, \( l \) is the length of the tube in the magnetic field, and \( B \) is the magnetic field's magnitude.

Step 2: Relating Force to Pressure Difference

Under static conditions, the Lorentz force \( F \) exerted by the magnetic field on the charged particles is balanced by a corresponding pressure force. This pressure force is \( \Delta p \times A \), where \( A \) is the cross-sectional area of the fluid, which can be expressed as \( w \times h \). Hence, \( \Delta p = \frac{F}{A} = \frac{J l B}{w h} = J l B \), assuming the forces balance and the mechanical force is converted to overcome the hydraulic resistance.

Step 3: Calculating Current Density for Given Pressure

Use the derived formula \( \Delta p = J l B \) to solve for \( J \). Given \( \Delta p = 1.00 \text{ atm} = 101325 \text{ Pa} \), \( B = 2.20 \text{ T} \), and \( l = 35.0 \text{ mm} = 0.035 \text{ m} \), substitute these values to find \( J = \frac{101325}{0.035 \times 2.20} \).

Step 4: Performing the Calculation

Following through with the calculation, \( J = \frac{101325}{0.035 \times 2.20} \approx 1314796 \text{ A/m}^2 \). This is the current density required to create a 1.00 atm pressure difference under the given conditions.

Key Concepts

Lorentz Force

Lorentz force is the fundamental principle driving the electromagnetic pump. It's a force that acts on charged particles moving within a magnetic field. When you have a charged particle with charge \( q \), moving with a velocity \( \mathbf{v} \) in a magnetic field \( \mathbf{B} \), the magnetic part of the Lorentz force can be described by the equation: \( F = q(\mathbf{v} \times \mathbf{B}) \). This means the direction of the force is perpendicular to both the velocity of the particle and the magnetic field.
In the context of the electromagnetic pump, we simplify this by applying it to a fluid where the bit of fluid moves perpendicular to the magnetic field. Here, the force occurs on charges due to the current density and provides a pressure difference to pump the fluid. The force per unit length on the fluid can be expressed as \( f = J l B \). This indicates that the force experienced by the fluid relies on:

• The current density \( J \), which shows how much current flows through a unit area
• The length \( l \) that the field encompasses
• The magnetic field magnitude \( B \)

This result shows how an electromagnetic pump uses Lorentz force to move conducting fluids efficiently.

Current Density

Current density refers to the amount of electrical current flowing per unit area through a cross-section of a conductor. This concept plays a unique role in the operation of an electromagnetic pump. In this scenario, current density \( J \) is crucial because it directly affects the Lorentz force, responsible for producing pressure differences within the fluid.
More specifically, current density \( J \) is used in the relationship \( J l B \), where \( l \) represents the length of the fluid exposed to the magnetic field, and \( B \) is the magnitude of the magnetic field. These aspects together determine the amount of force exerted on the fluid, hence the pressure difference created.
When tasked with creating a specific pressure difference, such as \( 1.00 \) atm in the exercise, you can rearrange the equation \( \Delta p = J l B \) to solve for \( J \). Knowing the desired pressure difference and other factors, you calculate the current density necessary to achieve the required force to pump the fluid:

• Finding \( J \) ensures enough force arises to achieve the needed pressure difference
• This calculated \( J \) should align with system limitations to avoid overheating or other inefficiencies

This illustrates just how integral current density is to finely tuning the operations of an electromagnetic pump.

Pressure Difference

In electromagnetic pumps, generating a pressure difference within a fluid is vital for effective pumping action. As per the scenario laid out in the exercise, this difference in pressure \( \Delta p \) arises when Lorentz force balances out the opposing pressure force, allowing for a smooth flow of the conducting fluid.
An important relationship exists: \( \Delta p = J l B \). This formula tells you how the pressure difference between two points in the fluid directly correlates with:

• The current density \( J \)
• The length \( l \) of the fluid in the magnetic field
• The magnetic field's strength \( B \)

To understand this further:

• Imagine \( \Delta p \) as the driving force pushing the fluid across different regions within the pump
• This force needs to be strong enough to counteract any resistance from the fluid, ensuring efficient operation

The exercise you encountered emphasizes calculating the pressure difference to ensure no fluid motion is disrupted by obstacles or by not having adequate force, further highlighting the elegance of using Lorentz force in the context of electromagnetic pumping.