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Chapter 30: Problem 31

What current is required in the windings of a long solenoid that has 1000 turns uniformly distributed over a length of \(0.400 \mathrm{m},\) to produce at the center of the solenoid a magnetic field of magnitude \(1.00 \times 10^{-4} \mathrm{T} ?\)

Short Answer

Expert verified
The required current in the windings of the solenoid to produce a magnetic field of the specified magnitude is 0.32 A.

Step by step solution

01

Determine the number of turns per unit length

First, calculate the number of turns per unit length (n) of the solenoid. This can be done by dividing the total number of turns (1000) by the total length (0.400 m). Thus \(n = \frac{1000}{0.400} = 2500 \, turns/meter\).
02

Isolate the current (I) in the formula

In Ampere's law, isolate the current. The formula is \(B = \mu_0 * (n * I)\). So, rearranging for \(I\), we get \(I = \frac{B}{\mu_0 * n}\).
03

Substitute the known values into the formula

Substituting \(B = 1.00 \times 10^{-4}T, \mu_0 = 4\pi \times 10^{-7} \, T*m/A, n = 2500 \, turns/meter\), we get \(I = \frac{1.00 \times 10^{-4}}{4\pi \times 10^{-7} \times 2500}A\). Calculating that gives \(I = 0.32 A\).

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Key Concepts

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

Ampere's Law
Ampere's Law is fundamental to understanding magnetic fields in conductors, including solenoids. It relates the magnetic field around a closed loop to the electric current passing through the loop. For a solenoid, which is essentially a coil wound in a helical shape, the magnetic field inside is nearly uniform and strongest at the center. The law is expressed as:
  • \( B = \mu_0 \cdot n \cdot I \)
where:
  • \( B \) is the magnetic field strength,
  • \( \mu_0 \) is the permeability of free space \((4\pi \times 10^{-7} \, T\cdot m/A)\),
  • \( n \) is the number of turns per unit length,
  • \( I \) is the current through the solenoid.
The law helps derive the current required to create a desired magnetic field inside a solenoid, like in the exercise provided. By using Ampere's Law and rearranging for \( I \), you can effectively find how much current is needed if the magnetic field and the turns per unit length are known.
Current Calculation
Calculating the required current in a solenoid involves rearranging Ampere's Law to solve for \( I \). This ensures that we can find the precise current needed to achieve a specific magnetic field inside the solenoid.
  • Rearrange the formula: \( I = \frac{B}{\mu_0 \cdot n} \)
  • Substitute the known values into the equation: \( B = 1.00 \times 10^{-4} T \), \( \mu_0 = 4\pi \times 10^{-7} \, T\cdot m/A \), and \( n = 2500 \, \text{turns/m} \).
  • Perform the calculation: \( I = \frac{1.00 \times 10^{-4}}{4\pi \times 10^{-7} \times 2500} = 0.32 \, A \)
This calculation will give you the exact current necessary to create the magnetic field specified, ensuring the solenoid functions as intended.
Turns per Unit Length
The turns per unit length, often denoted as \( n \), is an essential factor in determining the strength of the magnetic field in a solenoid. It reflects how tightly the coil is wound.
  • Calculate by dividing the total number of turns by the length of the solenoid: \( n = \frac{1000}{0.400} = 2500 \, \text{turns/m} \).
  • This value indicates how many turns exist in each meter of the solenoid.
  • The higher the \( n \), the stronger the magnetic field for a given current.
This calculation plays a critical role in using Ampere's Law, helping predict the behavior of the solenoid and ensuring its magnetic field meets specific requirements. Understanding this concept is key to designing solenoids for various applications.

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Most popular questions from this chapter

A toroid with a mean radius of \(20.0 \mathrm{cm}\) and 630 turns (see Fig. 30.30 ) is filled with powdered steel whose magnetic susceptibility \(\chi\) is \(100 .\) The current in the windings is 3.00 A. Find \(B\) (assumed uniform) inside the toroid.

The magnetic coils of a tokamak fusion reactor are in the shape of a toroid having an inner radius of \(0.700 \mathrm{m}\) and an outer radius of \(1.30 \mathrm{m} .\) The toroid has 900 turns of largediameter wire, each of which carries a current of \(14.0 \mathrm{kA}\) Find the magnitude of the magnetic field inside the toroid along (a) the inner radius and (b) the outer radius.

Three long wires (wire \(1,\) wire \(2,\) and wire 3 ) hang vertically. The distance between wire 1 and wire 2 is \(20.0 \mathrm{cm} .\) On the left, wire 1 carries an upward current of 1.50 A. To the right, wire 2 carries a downward current of \(4.00 \mathrm{A}\). Wire 3 is located such that when it carries a certain current, each wire experiences no net force. Find (a) the position of wire \(3,\) and (b) the magnitude and direction of the current in wire 3.

A circular coil of 5 turns and a diameter of \(30.0 \mathrm{cm}\) is oriented in a vertical plane with its axis perpendicular to the horizontal component of the Earth's magnetic field. A horizontal compass placed at the center of the coil is made to deflect \(45.0^{\circ}\) from magnetic north by a current of \(0.600 \mathrm{A}\) in the coil. (a) What is the horizontal component of the Earth's magnetic field? (b) The current in the coil is switched off. A "dip needle" is a magnetic compass mounted so that it can rotate in a vertical north-south plane. At this location a dip needle makes an angle of \(13.0^{\circ}\) from the vertical. What is the total magnitude of the Earth's magnetic field at this location?

Table \(P 30.70\) contains data taken for a ferromagnetic material. (a) Construct a magnetization curve from the data. Remember that \(\mathbf{B}=\mathbf{B}_{0}+\mu_{0} \mathbf{M} .\) (b) Determine the ratio \(B / B_{0}\) for each pair of values of \(B\) and \(B_{0},\) and construct a graph of \(B / B_{0}\) versus \(B_{0}\). (The fraction \(B / B_{0}\) is called the relative permeability, and it is a measure of the induced magnetic field.) $$\begin{array}{cc}B(T) & B_{0}(T) \\\\\hline 0.2 & 4.8 \times 10^{-5} \\\0.4 & 7.0 \times 10^{-5} \\\0.6 & 8.8 \times 10^{-5} \\\0.8 & 1.2 \times 10^{-4} \\\1.0 & 1.8 \times 10^{-4} \\\1.2 & 3.1 \times 10^{-4} \\\1.4 & 8.7 \times 10^{-4} \\\1.6 & 3.4 \times 10^{-3} \\\1.8 & 1.2 \times 10^{-1} \\\\\hline\end{array}$$
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