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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 1.40 \(\mathrm{m}\) long and 2.80 \(\mathrm{cm}\) in diameter. What current will you need to produce the necessary field?

Short Answer

Expert verified
The current needed is approximately 41.8 A.

Step by step solution

01

Understanding the Problem

We need to find the current required to produce a uniform magnetic field of 0.150 T in a solenoid. We know the solenoid is 1.40 m long, consists of 4000 turns, and has a diameter of 2.80 cm.
02

Formula for Magnetic Field in a Solenoid

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 \( (4\pi \times 10^{-7}\, \text{T m/A}) \), \( N \) is the number of turns, \( L \) is the length of the solenoid, and \( I \) is the current.
03

Rearranging the Formula to Find Current

To find the current \( I \), rearrange the formula: \( I = \frac{B \times L}{\mu_0 \times N} \).
04

Substitute the Known Values

Substitute \( B = 0.150 \) T, \( L = 1.40 \) m, \( N = 4000 \), and \( \mu_0 = 4\pi \times 10^{-7} \) Tm/A: \[ I = \frac{0.150 \times 1.40}{4\pi \times 10^{-7} \times 4000} \].
05

Solve for Current

Calculate the current \( I \): \[ I = \frac{0.150 \times 1.40}{4\pi \times 10^{-7} \times 4000} = \frac{0.21}{5.024 \times 10^{-3}} \approx 41.8 \text{ A} \].

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Key Concepts

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

Magnetic Field
Magnetic fields are invisible forces generated by electric currents. They play a crucial role in solenoid operation. Inside a solenoid, which is basically a coil of wire, a magnetic field is generated when an electric current passes through it. These fields can be quite uniform within the solenoid, particularly near the center, which is why solenoids are useful in a range of applications such as inductors or electromagnets.

In our exercise, the goal is to design a solenoid with a specific magnetic field strength, measured in teslas (T). To achieve a magnetic field of 0.150 T inside the solenoid, one must consider various factors such as the number of turns in the coil, the current flowing through the wire, and the length of the solenoid. The uniformity and strength of the field depend significantly on these parameters.

The formula used to calculate the magnetic field, given by \( B = \mu_0 \frac{N}{L} I \), highlights the direct relationship between the magnetic field \( B \) and the current \( I \), while also taking into account the number of turns \( N \) and solenoid length \( L \). By understanding how each of these variables influences the magnetic field, one can design solenoids to meet specific requirements.
Current Calculation
Calculating the current needed for a specific magnetic field in a solenoid involves some straightforward algebraic manipulation. Using the formula \( I = \frac{B \times L}{\mu_0 \times N} \), one can determine the amount of current, \( I \), necessary to achieve a desired field strength \( B \). This is achieved by rearranging the more commonly known formula for the magnetic field inside a solenoid.

In this solenoid design exercise, you begin with known values: a magnetic field \( B = 0.150 \) T, a solenoid length \( L = 1.40 \) m, and 4000 turns () of wire. The formula incorporates the permeability of free space \( \mu_0 = 4\pi \times 10^{-7} \) Tm/A. By substituting these values into the equation, one finds:
  • Magnetic Field \( B \): \( 0.150 \) T
  • Solenoid Length \( L \): \( 1.40 \) m
  • Number of Turns \( N \): \( 4000 \)
  • Permeability \( \mu_0 \): \( 4\pi \times 10^{-7} \) Tm/A
Calculating these, we find the required current is approximately 41.8 A. Understanding this calculation process is pivotal to designing solenoids that function as intended to produce the necessary magnetic fields without exceeding material capabilities or design limitations.
Circuits
A solenoid is a critical component in electrical circuits, especially when a magnetic field is desired. Comprising numerous turns of wire, a solenoid operates within a circuit to convert electrical energy into magnetic energy.

In a circuit, the solenoid connects to a power source that supplies direct current. This current flows through the wire coil, generating a magnetic field. In our design scenario, the solenoid's configuration—defined by its length, diameter, and the number of turns—dictates how effectively it can produce the desired magnetic field with a given current.

To integrate a solenoid into a circuit effectively:
  • Ensure the power supply provides sufficient amplitude (current) to achieve the target magnetic field strength.
  • Consider resistance and heat generation over long periods of operation, as heating can lead to changes in the solenoid's physical properties or insulation breakdown.
  • Account for the solenoid’s inductance, which could affect circuit timing if the solenoid is part of an alternating current (AC) system.
Overall, understanding the interplay between the solenoid and other circuit components is essential for the system’s performance, ensuring that the solenoid operates efficiently and safely within its design specifications.

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Most popular questions from this chapter

Fields Within the Atom. In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{m}\) with a speed of \(2.2 \times 10^{6} \mathrm{m} / \mathrm{s} .\) If we are viewing the atom in such a way that the electron's orbit is in the plane of the paper with the electron moving clockwise, find the magnitude and direction of the electric and magnetic fields that the electron produces at the location of the nucleus (treated as a point).

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