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Chapter 24: Problem 13

A solenoid used to produce magnetic fields for research purposes is \(2.0 \mathrm{m}\) long, with an inner radius of \(30 \mathrm{cm}\) and 1000 turns of wire. When running, the solenoid produces a field of \(1.0 \mathrm{T}\) in the center. Given this, how large a current does it carry?

Short Answer

Expert verified
The current that the solenoid carries is approximately \(159\, \mathrm{A}\).

Step by step solution

01

Determine the Given Variables

First, identify all of the given variables. The magnetic field \(B\) in the centre of the solenoid is given as \(1.0 \, \mathrm{T}\). The number of turns of wire (N) in the solenoid is 1000. The length of the solenoid (L) is \(2.0\, \mathrm{m}\).
02

Determine Turns per Unit Length

Here, we calculate the number of turns per unit length (n). We do this by dividing the total number of turns(N) by the length of the solenoid (L). \[n = \frac{N}{L} = \frac{1000}{2}\] which results in \(n = 500\, \mathrm{turns/m}\)
03

Rearrange the Magnetic Field Equation and Compute for Current

We'll use the solenoid's magnetic field formula: \(B = \mu_{0} * n * i\). Given the magnetic field \(B\), vacuum permeability \(\mu_{0} = 4\pi x 10^{-7}\, \mathrm{Tm/A}\), and number of turns per unit length \(n\), we can solve for current \(i\). Rearranging the equation gives \[i = \frac{B}{\mu_{0}n} = \frac{1}{4\pi x 10^{-7}*500}\]
04

Evaluate and Round the Result

Performing the calculation yields \(i \approx 159.2\, \mathrm{A}\). In accordance with significant figures rules, we round this to three figures. Therefore the required current is approximately \(159\, \mathrm{A}\).

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Key Concepts

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

Magnetic Field
The concept of a magnetic field is central to understanding how current in a solenoid gives rise to a magnetic phenomenon. A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. It exerts a force known as the Lorentz force, which can cause materials like iron to align along the direction of the field, obscure items like compass needles to point north, and charged particles to move in circular paths.

In the case of a solenoid鈥攁 coil of wire designed to produce a magnetic field when electric current passes through it鈥攖he magnetic field is strongest inside the coil, along the central longitudinal axis. This field can be visualized as a series of concentric circles that loop around the wire of the solenoid, perpendicular to the current flow.

In the given exercise, the magnetic field strength in the center of the solenoid is provided as 1.0 Tesla (T), a unit of measurement for magnetic field strength, indicating a relatively strong field suitable for research applications.
Turns per Unit Length
In electromagnetism, particularly when analyzing solenoids, 'turns per unit length' is a crucial concept. It signifies the number of windings of the coil, or turns of the wire, per unit length of the solenoid. The density of the windings directly influences the strength of the magnetic field generated by a current running through the solenoid. Higher density (more turns per unit length) results in a stronger magnetic field, given a constant current.

To calculate the 'turns per unit length' (), you divide the total number of turns (N) by the length (L) of the solenoid. For example, a solenoid with 1000 turns of wire that is 2 meters long would have a turns density of 500 turns per meter. This figure is essential for determining the strength of the magnetic field created by the solenoid, as seen in the exercise where a critical step involves using this information to calculate the current powering the solenoid.
Ampere's Law
Ampere's Law is one of the fundamental equations of electromagnetism, relating the integrated magnetic field around a closed loop to the electric current passing through the loop. It is part of Maxwell's equations, which are the basis of classical electromagnetism. Specifically, Ampere's Law states that the circular magnetic field around a current-carrying conductor is proportional to the current in the conductor, with the constant of proportionality being the permeability of free space (mu_{0}).

When dealing with a solenoid, Ampere's Law simplifies to the formula for the magnetic field inside a solenoid: B = mu_{0} * n * i, where B is the magnetic field, mu_{0} the magnetic constant or the permeability of free space, the number of turns per unit length, and i the current through the solenoid.

This relationship enables us to solve for the current when given the magnetic field and the number of turns per unit length, as demonstrated in the step-by-step solution to the exercise where the current needed to create a field of 1.0 T was found to be approximately 159 A.

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Most popular questions from this chapter

All ferromagnetic materials have a Curie temperature, a temperature above which they will cease to be magnetic. Explain in some detail why you might expect this to be so.

In a simplified model of the hydrogen atom, its electron moves in a circular orbit with a radius of \(5.3 \times 10^{-10} \mathrm{m}\) at a frequency of \(6.6 \times 10^{15}\) Hz. What magnetic field would be required to cause an electron to undergo this same motion?

II A square current loop \(5.0 \mathrm{cm}\) on each side carries a \(500 \mathrm{mA}\) current. The loop is in a \(1.2 \mathrm{T}\) uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is \(30^{\circ}\) away from the field direction. What is the magnitude of the torque on the current loop?

Two concentric current loops lie in the same plane. The smaller loop has a radius of \(3.0 \mathrm{cm}\) and a current of \(12 \mathrm{A}\). The bigger loop has a current of 20 A. The magnetic field at the center of the loops is found to be zero. What is the radius of the bigger loop?

A loudspeaker creates sound by pushing air back and forth with a paper cone that is driven by a magnetic force on a wire coil at the base of the cone. Figure \(\mathrm{P} 24.52\) shows the details. The cone is attached to a coil of wire that sits in the gap between the poles of a circular magnet. The 0.18 T magnetic field, which points radially outward from \(\mathrm{N}\) to \(\mathrm{S}\), exerts a force on a current in the wire,moving the cone. The coil of wire that sits in this gap has a diameter of \(5.0 \mathrm{cm},\) contains 20 turns of wire, and has a resistance of \(8.0 \Omega .\) The speaker is connected to an amplifier whose instantaneous output voltage of \(6.0 \mathrm{V}\) creates a clockwise current in the coil as seen from above. What is the magnetic force on the coil at this instant?
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