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Chapter 28: Problem 26

(I) A 32 -cm-long solenoid, \(1.8 \mathrm{~cm}\) in diameter, is to produce a 0.30-T magnetic field at its center. If the maximum current is \(4.5 \mathrm{~A}\), how many turns must the solenoid have?

Short Answer

Expert verified
The solenoid must have approximately 16,977 turns.

Step by step solution

01

Understand the Formula for Magnetic Field in a Solenoid

The magnetic field inside a solenoid is given by the formula \( B = \mu_0 \cdot n \cdot I \), where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space \( (\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot \text{m/A}) \), \( n \) is the number of turns per unit length, and \( I \) is the current.
02

Express Number of Turns Per Unit Length

Given \( n = \frac{N}{L} \), where \( N \) is the total number of turns and \( L \) is the length of the solenoid, we can express \( n \) in terms of \( N \):\[ n = \frac{N}{0.32} \text{ m} \] since the length of the solenoid is 32 cm (0.32 m).
03

Simplify and Insert Known Values Into Formula

We need to rearrange the formula to solve for \( N \). Substitute \( n = \frac{N}{0.32} \) and \( I = 4.5 \text{ A} \) into the magnetic field formula:\[ 0.30 = (4\pi \times 10^{-7}) \cdot \left(\frac{N}{0.32}\right) \cdot 4.5 \]
04

Solve for Total Number of Turns (N)

Rearrange the equation to solve for \( N \):\[ N = \frac{0.30 \times 0.32}{4\pi \times 10^{-7} \times 4.5} \]First calculate the denominator:\[ 4\pi \times 10^{-7} \times 4.5 = 5.65487 \times 10^{-6} \]Now solve for \( N \):\[ N = \frac{0.096}{5.65487 \times 10^{-6}} \approx 16976.61 \]Therefore, the number of turns required is approximately 16977 turns.

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Key Concepts

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

Solenoid
A solenoid is a coil of wire that is wound in a helix shape, resembling a cylindrical form. When an electric current flows through a solenoid, it generates a magnetic field. This property is leveraged in many applications, ranging from electromagnets to inductors in electrical circuits. To gain a better understanding of how a solenoid works, consider imagining a tightly coiled spring. When carrying a current, the solenoid behaves like a bar magnet with a north and a south pole, and the magnetic field lines run parallel to the coil, following its helical path. This configuration is crucial in understanding how solenoids can be used to create controlled magnetic fields.
  • Applications include switches, relays, and motors.
  • The magnetic field strength depends on the current and the number of turns.
Understanding these basic properties sets the stage for exploring how solenoids can be designed to produce specific magnetic field strengths.
Current in a Solenoid
The current flowing through a solenoid is a major factor in determining the strength of the magnetic field it produces. The relationship between the current and the magnetic field in a solenoid is linear, as given by the formula:\[ B = \mu_0 \cdot n \cdot I \]Where:
  • \( B \) is the magnetic field strength inside the solenoid.
  • \( \mu_0 \) is the permeability of free space, a constant \( (4\pi \times 10^{-7} \text{ T}\cdot \text{m/A}) \).
  • \( n \) is the number of turns per unit length of the solenoid.
  • \( I \) is the current through the solenoid.
Increasing the current in the solenoid will therefore directly increase the magnetic field strength. However, practical limits come into play, such as the maximum current the wire can handle without overheating or causing damage. Balancing these factors is crucial in the safe operation of devices that use solenoids.
Number of Turns in a Solenoid
The number of turns in a solenoid, denoted as \( N \), plays a crucial role in determining the magnetic field it produces. More turns lead to a stronger magnetic field, assuming other factors like current remain constant. We can define the number of turns per unit length as \( n = \frac{N}{L} \), where \( L \) is the length of the solenoid.To find \( N \), we can rearrange the formula for the magnetic field:\[ B = \mu_0 \cdot \left(\frac{N}{L}\right) \cdot I \]Rearranging to solve for \( N \):\[ N = \frac{B \cdot L}{\mu_0 \cdot I} \]In the example provided, a solenoid needs to create a magnetic field of 0.30 T with a current of 4.5 A over a length of 0.32 meters. By substituting these values into the equation, one can solve for the number of turns needed. In this case, approximately 16,977 turns are required.
  • More turns per unit length increases both cost and complexity.
  • Each application may require a different configuration depending on the magnetic field needed.
Understanding how the number of turns affects the magnetic field helps in designing solenoids for various technological and industrial applications.

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Most popular questions from this chapter

(1I) A single point charge \(q\) is moving with velocity \(\vec{v}\) . Use the Biot-Savart law to show that the magnetic field \(\vec{\mathbf{B}}\) it produces at a point \(P\) , whose position vector relative to the charge is \(\vec{\mathbf{r}}(\) Fig. 46\(),\) is given by $$\vec{\mathbf{B}}=\frac{\mu_{0}}{4 \pi} \frac{q \vec{\mathbf{v}} \times \vec{\mathbf{r}}}{r^{3}}$$

(II) Two long thin parallel wires 13.0 \(\mathrm{cm}\) apart carry \(35-\mathrm{A}\) currents in the same direction. Determine the magnetic field vector at a point 10.0 \(\mathrm{cm}\) from one wire and 6.0 \(\mathrm{cm}\) from the other

(II) A 20.0 -m-long copper wire, \(2.00 \mathrm{~mm}\) in diameter including insulation, is tightly wrapped in a single layer with adjacent coils touching, to form a solenoid of diameter \(2.50 \mathrm{~cm}\) (outer edge). What is \((a)\) the length of the solenoid and \((b)\) the field at the center when the current in the wire is \(16.7 \mathrm{~A}\) ?

(II) A compass needle points \(28^{\circ}\) E of \(N\) outdoors. However, when it is placed 12.0 \(\mathrm{cm}\) to the east of a vertical wire inside a building, it points \(55^{\circ} \mathrm{E}\) of \(\mathrm{N} .\) What is the magnitude and direction of the current in the wire? The Earth's field there is \(0.50 \times 10^{-4} \mathrm{T}\) and is horizontal.

3\. (1II) A very long flat conducting strip of width \(d\) and negli- gible thickness lies in a horizontal plane and carries a uniform current \(I\) across its cross section. \((a)\) Show that at points a distance \(y\) directly above its center, the field is given by $$B=\frac{\mu_{0} I}{\pi d} \tan ^{-1} \frac{d}{2 y}$$ assuming the strip is infinitely long. [Hint. Divide the strip into many thin wires and sum (integrate) over these. (b) What value does \(B\) approach for \(y \gg d ?\) Does this make sense? Explain.
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