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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 \(55.0 \mathrm{~cm}\) 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 to produce the necessary magnetic field is calculated to be approximately 1.59 Amperes.

Step by step solution

01

Identify given quantities

The problem provides the following quantities: magnetic field B = 0.150 T, turns of wire = 4000, length of the solenoid = 55.0 cm, diameter of the solenoid (not needed for this problem) = 2.80 cm. We need to find the current I.
02

Convert units

First, the length of the solenoid needs to be converted into meters. So, length = \(55.0 \times 10^{-2}\) m = 0.55 m.
03

Calculate number of turns per unit length (n)

The total number of turns is 4000. Therefore, the number of turns per unit length (n) is the total number of turns divided by the length of the solenoid. Substituting the known values gives \(n = \frac{4000}{0.55}\) turns per meter.
04

Use Ampere's law to calculate the current

Rearrange Ampere's law to solve for I: \(I = \frac{B}{\mu_0 * n}\). Substituting the known values, including for \(\mu_0 = 4\pi \times 10^{-7} Tm/A\), gives \(I = \frac{0.150}{4\pi \times 10^{-7} \times \frac{4000}{0.55}}\) Ampere.

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Key Concepts

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

Ampere's Law
To effectively grasp the calculations surrounding solenoids and magnetic fields, it's critical to comprehend Ampere's Law. This law is a fundamental principle in electromagnetism that's central to understanding how current creates a magnetic field. Ampere's Law states that the integral of the magnetic field (B) along a closed loop is proportional to the electric current (I) passing through the loop. The law is conventionally expressed as \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I \), where \( \mu_0 \) is the permeability of free space, a physical constant. In simpler terms, Ampere's Law connects the magnetism generated by an electric current with the strength and the direction of the magnetic field.

This principle allows us to calculate the magnetic field inside a solenoid – a long, coiled wire through which a current passes. Given the symmetrical nature of a solenoid, Ampere's Law simplifies to \( B = \mu_0 \cdot n \cdot I \), where \( n \) is the number of turns per unit length of the solenoid, and \( B \) is the magnetic field strength at the solenoid's center. This equation is crucial as it provides a direct method to calculate the current needed for a desired magnetic field, which is the essence of the task at hand.
Magnetic Field Strength
Magnetic field strength, often symbolized as \( B \), is a quantitative measure of the magnetism of a magnetic field. It is defined as the force exerted per unit length on a current-carrying wire placed at right angles to the field. The unit of magnetic field strength is the Tesla (T), with one Tesla being equivalent to one Newton per ampere per meter. In the situation of a solenoid, the magnetic field is generated due to electric current running through the coils of wire.

The strength of this field is highly uniform in the central region of a long solenoid and is directly proportional to the current \( I \) and the number of turns per unit length \( n \) inside the solenoid. This is represented by the equation \( B = \mu_0 n I \), where \( \mu_0 \) equals \( 4\pi \times 10^{-7} \) Tm/A, known as the magnetic constant. The strength diminishes rapidly outside the solenoid ends, so calculations often focus on the field within the central part. This concept is critical because it helps predict the behavior of the solenoid under different current magnitudes and provides a direct way to engineer the desired magnetic field by adjusting the current flowing through the solenoid.
Turns Per Unit Length
The concept of 'turns per unit length' in the context of a solenoid is quite straightforward but essential for calculating magnetic fields. It denotes the number of coils of wire per unit length of the solenoid. Mathematically, it is expressed as \( n = \frac{N}{l} \), where \( n \) is the number of turns per unit length, \( N \) is the total number of turns, and \( l \) is the length of the solenoid. By increasing the number of turns per unit length, one can enhance the magnetic field strength for a given current.

In our exercise, the total number of turns is 4000, and the solenoid is 0.55 meters long, resulting in \( n = \frac{4000}{0.55} \) turns per meter. This figure is vital for calculating the current needed, as it directly affects the magnetic field strength. The density of the coils means that within every meter of the solenoid's length are many loops through which the current circulates, each contributing to the overall magnetic field. Properly understanding and calculating 'turns per unit length' ensures precision in designing a solenoid capable of producing the exact magnetic field required for an application.

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Most popular questions from this chapter

A long, straight wire lies along the \(x\) -axis and carries current \(I_{1}=2.00 \mathrm{~A}\) in the \(+x\) -direction. A second wire lies in the \(x y\) -plane and is parallel to the \(x\) -axis at \(y=+0.800 \mathrm{~m}\). It carries current \(I_{2}=6.00 \mathrm{~A}\), also in the \(+x\) -direction. In addition to \(y \rightarrow \pm \infty,\) at what point on the \(y\) -axis is the resultant magnetic field of the two wires equal to zero?

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 \(J=\alpha r,\) where \(\alpha\) is a constant. (a) \(\mathrm{By}\) the requirement that \(J\) integrated over the cross section of the wire gives the total current \(I,\) calculate the constant \(\alpha\) 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 \(I\).

A solenoid is designed to produce a magnetic field of \(0.0270 \mathrm{~T}\) at its center. It has radius \(1.40 \mathrm{~cm}\) and length \(40.0 \mathrm{~cm},\) and the wire can carry a maximum current of 12.0 A. (a) What minimum number of turns per unit length must the solenoid have? (b) What total length of wire is required?

The magnetic field around the head has been measured to be approximately \(3.0 \times 10^{-8} \mathrm{G}\). Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modeling them as a single circular current loop \(16 \mathrm{~cm}\) (the width of a typical head) in diameter. What is the current needed to produce such a field at the center of the loop?

Long, straight conductors with square cross sections and each carrying current \(I\) are laid side by side to form an infinite current sheet (Fig. \(\mathbf{P 2 8 . 6 9}\) ). The conductors lie in the \(x y\) -plane, are parallel to the \(y\) -axis, and carry current in the \(+y\) -direction. There are \(n\) conductors per unit length measured along the \(x\) -axis. (a) What are the magnitude and direction of the magnetic field a distance \(a\) below the current sheet? (b) What are the magnitude and direction of the magnetic field a distance \(a\) above the current sheet?

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