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Chapter 28: Problem 44

A toroidal solenoid (see Example 28.10 ) has inner radius \(r_{1}=15.0 \mathrm{cm}\) and ourer radius \(r_{2}=18.0 \mathrm{cm} .\) The solenoid has 250 turns and carries a current of 8.50 A. What is the magnitude of the magnetic field at the following distances from the center of the torus: (a) \(12.0 \mathrm{cm} ;(\mathrm{b}) 16.0 \mathrm{cm} ;(\mathrm{c}) 20.0 \mathrm{cm} ?\)

Short Answer

Expert verified
(a) 0 T; (b) 4.43 mT; (c) 0 T.

Step by step solution

01

Understand the Problem

We need to find the magnitude of the magnetic field at different distances from the center of a toroidal solenoid. The toroidal solenoid has an inner radius of 15.0 cm, an outer radius of 18.0 cm, 250 turns, and carries a current of 8.50 A.
02

Identify Key Formula

Inside a toroidal solenoid, the magnetic field at a distance \( r \) from the center is given by the formula: \( B = \frac{\mu_0 n I}{2 \pi r} \), where \( n \) is the number of turns per unit length, \( I \) is the current, and \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}) \). Here, \( n = \frac{N}{2\pi (r_2-r_1)} \), where \( N \) is the total number of turns.
03

Calculate Magnetic Field for Part (a)

For distance \( r = 12.0 \text{ cm} = 0.12 \text{ m} \), it is below the inner radius, therefore there is no magnetic field inside the toroidal solenoid. Thus, \( B = 0 \text{ T} \).
04

Calculate Magnetic Field for Part (b)

For distance \( r = 16.0 \text{ cm} = 0.16 \text{ m} \), which lies between inner and outer radii:1. Calculate \( n = \frac{250}{2\pi (0.18 - 0.15)} \approx \frac{250}{2\pi \times 0.03} \approx 1326.914 \, \text{turns/m} \).2. Use the formula: \[ B = \frac{4\pi \times 10^{-7} \times 1326.914 \times 8.5}{2 \pi \times 0.16} \approx 4.43 \times 10^{-3} \, \text{T} \approx 4.43 \, \text{mT} \].
05

Calculate Magnetic Field for Part (c)

For distance \( r = 20.0 \text{ cm} = 0.20 \text{ m} \), which is outside the toroidal solenoid, the magnetic field is zero, as there is no magnetic field outside the solenoid. Thus, \( B = 0 \text{ T} \).

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Key Concepts

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

Magnetic Field Calculation
The magnetic field generated by a toroidal solenoid, such as the one described in this exercise, is a function of several key parameters. Understanding how to calculate this field is essential in electromagnetism.
To calculate the magnetic field at a point, the distance from the solenoid's center, or radius, is crucial. The formula for the magnetic field inside a toroidal solenoid is given by:
  • \[ B = \frac{\mu_0 n I}{2 \pi r} \]
Here, \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, \(I\) is the current passing through the solenoid, and \(r\) is the radial distance from the center of the torus.
This formula reveals that the magnetic field is inversely proportional to the radius, meaning the field strength decreases as you move away from the center of the solenoid.
Ampere's Law
Ampere's Law is a fundamental principle of electromagnetism that provides significant insight when solving problems related to magnetic fields. It relates the integrated magnetic field around a closed loop to the electric current passing through the loop.
The general form of Ampere's Law can be expressed as:
  • \[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc} \]
In this equation, the left side is the line integral of the magnetic field \(\mathbf{B}\) along a closed loop, and \(I_{enc}\) denotes the current enclosed by the loop.
When applied to a toroidal solenoid, the law simplifies the analysis of the magnetic field inside the solenoid itself, allowing the derivation of the practical formula used to compute \(B\) for any point within its bounds.
Electromagnetism
Electromagnetism is a branch of physics that studies the interactions between electric charges and currents through magnetic fields. It combines two essential concepts: electricity and magnetism.
In the context of a toroidal solenoid, electromagnetism helps us understand how a current-carrying coil generates a magnetic field. The current passing through the solenoid creates concentric loops of magnetic field lines within the solenoid, impacting points both within the coil and beyond its physical boundaries.
Essentially, electromagnetism describes how the movement of electric charges (current) through a conductor (like a solenoid) results in the generation of magnetic forces and fields, embodying the dynamic relationship between these two phenomena.

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Most popular questions from this chapter

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 \(\mathrm{A}\) . (a) What minimum number of turns per unit length must the solenoid have? (b) What total length of wire is required?

A toroidal solenoid with 500 turns is wound on a ring with a mean radins of 290 \(\mathrm{cm}\) . Find the current in the winding that is required to set up a magnetic field of 0.350 \(\mathrm{T}\) in the ring (a) if the ring is made of annealed iron \(\left(K_{m}=1400\right)\) and \((b)\) if the ring is made of sillicon steel \(\left(K_{m}=5200\right)\) .

A very long, straight horizontal wire carries a current such that \(3.50 \times 10^{18}\) 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?

A neophyte magnet designer tells you that he can produce a magnetic field \(\vec{B}\) in vacuum that points everywhere in the \(x\) -direction and that increases in magnitude with increasing \(x\) . That is, \(\vec{B}=B_{0}(x / a) \hat{\imath},\) where \(B_{0}\) and \(a\) are constants with units of teslas and meters, respectively. Use Gauss's law for magnetic fields to show that this claim is impossible. (Hint: Use a Gaussian surface in the shape of a rectangular box, with edges parallel to the \(x\) ; \(y\) . and \(z\) -axes.)

A long, straight, solid cylinder, oriented with its axis in the z-direction, carries a current whose current density is \(\overrightarrow{\boldsymbol{J}}\) . The current density, although symmetrical about the cylinder axis, is not constant and varies according to the relationship $$ \begin{array}{rlrl}{\overrightarrow{\boldsymbol{J}}} & {=\left(\frac{b}{r}\right) e^{(r-a) / \delta} \hat{\boldsymbol{k}}} & {} & {\text { for } r \leq a} \\ {} & {=\mathbf{0}} & {} & {\text { for } \boldsymbol{r} \geq a}\end{array} $$ where the radius of the cylinder is \(a=5.00 \mathrm{cm}, r\) is the radial distance from the cylinder axis, \(b\) is a constant equal to \(600 \mathrm{A} / \mathrm{m},\) and \(\delta\) is a constant equal to \(2.50 \mathrm{cm} .\) (a) Let \(I_{0}\) be the total current passing through the entire cross section of the wire. Obtain an expression for \(I_{0}\) in terms of \(b, \delta,\) and \(a .\) Evaluate your expression to obtain a numerical value for \(I_{0}\) . \((b)\) Using Ampere's law, derive an expression for the magnetic field \(\vec{B}\) in the region \(r \geq a\) . Express your answer in terms of \(I_{0}\) rather than \(b\) . (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. Express your answer in terms of \(I_{0}\) rather than \(b\) . (d) Using Ampere's law, derive an expression for the magnetic field \(\vec{B}\) in the region \(r \leq a\) (e) Evaluate the magnitude of the magnetic field at \(r=\delta, r=a,\) and \(r=2 a\) .
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