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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 closely wound, circular coil with radius \(2.40 \mathrm{~cm}\) has 800 turns. (a) What must the current in the coil be if the magnetic field at the center of the coil is \(0.0770 \mathrm{~T}\) ? (b) At what distance \(x\) from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?

A \(-4.80 \mu \mathrm{C}\) charge is moving at a constant speed of \(6.80 \times 10^{5} \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction relative to a reference frame. At the instant when the point charge is at the origin, what is the magneticfield vector it produces at the following points: (a) \(x=0.500 \mathrm{~m}, y=0\) \(z=0 ;\) (b) \(x=0, y=0.500 \mathrm{~m}, z=0 ;\) (c) \(x=0.500 \mathrm{~m}, y=0.500 \mathrm{~m}\) \(z=0 ;(\mathrm{d}) x=0, y=0, z=0.500 \mathrm{~m} ?\)

A magnetic field of \(37.2 \mathrm{~T}\) has been achieved at the MIT Francis Bitter Magnet Laboratory. Find the current needed to achieve such a field (a) \(2.00 \mathrm{~cm}\) from a long, straight wire; (b) at the center of a circular coil of radius \(42.0 \mathrm{~cm}\) that has 100 turns; (c) near the center of a solenoid with radius \(2.40 \mathrm{~cm},\) length \(32.0 \mathrm{~cm},\) and 40,000 turns.

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).

A very long, straight horizontal wire carries a current such that \(8.20 \times 10^{18}\) electrons per second pass any given point going from west to east. What are the magnitude and direction of the magnetic field this wire produces at a point \(4.00 \mathrm{~cm}\) directly above it?

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