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Chapter 28: Problem 42

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150-T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 55.0 cm long and 2.80 cm in diameter. What current will you need to produce the necessary field?

Short Answer

Expert verified
The current needed is approximately 16.23 A.

Step by step solution

01

Understanding the Problem

We need to find the current required to create a magnetic field of 0.150 T inside a solenoid. The solenoid has 4000 turns, is 55.0 cm long, and has a diameter of 2.80 cm.
02

Identify the Solenoid Formula

The magnetic field inside a solenoid is given by the formula \( B = \mu_0 \frac{N}{L} I \), where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \text{ Tm/A}) \), \( N \) is the number of turns, \( L \) is the length of the solenoid in meters, and \( I \) is the current in amperes.
03

Convert Solenoid Length to Meters

The length of the solenoid is given as 55.0 cm. Convert this to meters: \( 55.0 \text{ cm} = 0.550 \text{ m} \).
04

Substitute Known Values into the Formula

Substitute \( B = 0.150 \text{ T} \), \( N = 4000 \), \( L = 0.550 \text{ m} \), and \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \) into the magnetic field formula: \( 0.150 = (4\pi \times 10^{-7}) \frac{4000}{0.550} I \).
05

Solve for Current \( I \)

Rearrange the equation to solve for \( I \): \( I = \frac{0.150 \times 0.550}{4\pi \times 10^{-7} \times 4000} \). Simplify this expression to find \( I \).
06

Calculate the Current

Perform the calculation: \( I = \frac{0.150 \times 0.550}{4\pi \times 10^{-7} \times 4000} \approx 16.23 \text{ A} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field Calculation
Calculating the magnetic field of a solenoid is essential for understanding how solenoids work in practice. Solenoids are coils of wire that generate a magnetic field when an electric current flows through them. The formula used to calculate the magnetic field inside a solenoid is given by \[ B = \mu_0 \frac{N}{L} I \]where:
  • \( B \) is the magnetic field in teslas (T).
  • \( \mu_0 \) is the permeability of free space, a constant valued at \( 4\pi \times 10^{-7} \text{ Tm/A} \).
  • \( N \) represents the number of turns of the wire.
  • \( L \) is the length of the solenoid in meters.
  • \( I \) is the electrical current in amperes (A).
Knowing these variables allows you to calculate the strength of the magnetic field that the solenoid will produce. The equation shows that the magnetic field is directly proportional to the number of turns and the current, and inversely proportional to the length of the solenoid. This understanding is crucial for designing solenoids in various applications.
Permeability of Free Space
The permeability of free space, denoted as \( \mu_0 \), is a fundamental constant in physics that helps describe how magnetic fields interact with the vacuum of space. Its precise value is \( 4\pi \times 10^{-7} \text{ Tm/A} \). This constant is vital in the magnetic field calculation of a solenoid because it provides a standard by which the strength of the magnetic field can be measured without the influence of external materials.

It's fundamental in our equation \( B = \mu_0 \frac{N}{L} I \), as it relates the number of wire turns and the current to the magnetic field. One of the elegant properties of \( \mu_0 \) is that it helps maintain the proportional relationship between these quantities in a vacuum. Understanding \( \mu_0 \) is important for anyone working with solenoids or any system based on electromagnetism, as it shows how magnetic fields behave in the absence of conductive or magnetic materials.
Solenoid Design
Designing a solenoid requires carefully considering several physical parameters to achieve the desired magnetic field strength. For engineering purposes, this includes the total number of turns \( (N) \), the length \( (L) \), and the diameter of the solenoid, which affects the physical size but not the magnetic field directly.

The solenoid in the exercise has a length of 55.0 cm, which must be converted to meters (0.550 m) for use in formulas. Its diameter is 2.80 cm, which informs the wire arrangement for the turns. Meanwhile, the number of turns (4000) dictates the tightness of the coil.

A well-designed solenoid should produce a uniform magnetic field at its center, which is crucial for applications in magnetic resonance imaging (MRI) machines, electromagnets, and inductors in circuits. In our scenario, achieving the necessary magnetic field of 0.150 T depends on how these parameters are chosen and combined with the electric current through the wire. Such principles guide the design in practical applications where precision in magnetic field strength is crucial.

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Most popular questions from this chapter

A long, straight wire with a circular cross section of radius \(R\) carries a current \(I\). Assume that the current density is not constant across the cross section of the wire, but rather varies as \(J =\) \(ar\), where a is a constant. (a) By the requirement that \(J\) integrated over the cross section of the wire gives the total current \(I\), calculate the constant \(a\) in terms of \(I\) and \(R\). (b) Use Ampere's law to calculate the magnetic field \(B(r)\) for (i) \(r\) \(\leq\) R and (ii) \(r\) \(\geq\) R. Express your answers in terms of \(I\).

Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percentage of the earth's magnetic field, which is 0.50 G. Is this value cause for worry?

A long, straight, solid cylinder, oriented with its axis in the \(z\)-direction, carries a current whose current density is \(\overrightarrow{J}\). The current density, although symmetric about the cylinder axis, is not constant and varies according to the relationship $$\overrightarrow{J} = (\frac{b}{r})e^{(r a)/\delta}\hat{k} \space for \space r \leq a$$ $$=0 \space for \space r \geq a$$ where the radius of the cylinder is a = 5.00 cm, \(r\) is the radial distance from the cylinder axis, \(b\) is a constant equal to 600 A/m, and \(\delta\) is a constant equal to 2.50 cm. (a) Let \(I_0\) be the total current passing through the entire cross section of the wire. Obtain an expression for \(I_0\) in terms of \(b\), \(\delta\), and a. Evaluate your expression to obtain a numerical value for I0. (b) Using Ampere's law, derive an expression for the magnetic field \(\overrightarrow{B}\) in the region \(r \leq a\). Express your answer in terms of \(I_0\) rather than b. (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. Express your answer in terms of \(I_0\) rather than b. (d) Using Ampere's law, derive an expression for the magnetic field \(\overrightarrow{B}\) in the region \(r \leq a\). (e) Evaluate the magnitude of the magnetic field at \(r = \delta\), \(r = a\), and \(r = 2a\).

The current in the windings of a toroidal solenoid is 2.400 A. There are 500 turns, and the mean radius is 25.00 cm. The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be 1.940 T. Calculate (a) the relative permeability and (b) the magnetic susceptibility of the material that fills the toroid.

A 15.0-cm-long solenoid with radius 0.750 cm is closely wound with 600 turns of wire. The current in the windings is 8.00 A. Compute the magnetic field at a point near the center of the solenoid.
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