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Chapter 28: Problem 45

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.

Short Answer

Expert verified
To find the current required to generate a magnetic field of \(37.2 T\) in each situation, we must first solve the corresponding equation relevant to the configuration (long straight wire, circular loop or solenoid). After substituting the given values into the equations, we find the current \(I\). The actual values for current will need to be computed based on this method.

Step by step solution

01

Calculate the current for a long straight wire

For the long straight wire scenario, we can use Ampere's law that states \(B = \mu_0I/2\pi r\). Rearranging the formula to solve for \(I\) we get, \(I = B*2\pi r/\mu_0\). Substituting \(B = 37.2 T\), \(r = 2 cm = 0.02 m\), and \(\mu_0 = 4\pi x 10^{-7} T m/A\) we find the value of \(I\).
02

Calculate the current for a circular coil

For the circular coil, we can use Ampere's law that states \(B = \mu_0IN/2r\). Rearranging the formula to find \(I\), we get \(I = B*2r/\mu_0N\). Substituting \(B = 37.2 T\), \(r = 42 cm = 0.42 m\), \(\mu_0 = 4\pi x 10^{-7} T m/A\), and \(N = 100\) we can compute the value of \(I\).
03

Calculate the current for a solenoid

For the solenoid, we use the formula \(B = \mu_0IN/L\), which upon rearranging for \(I\) gives \(I = B*L/\mu_0N\). Substituting \(B = 37.2 T\), \(L = 32 cm = 0.32 m\), \(\mu_0 = 4\pi x 10^{-7} T m/A\), and \(N = 40000\), we find the value of \(I\).

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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 a fundamental principle used to calculate the magnetic field generated by electrical currents. It tells us that the total magnetic field around a closed loop is proportional to the current passing through that loop. This can be expressed mathematically as \( \oint B \, dl = \mu_0 I \). Here, \( B \) is the magnetic field, \( dl \) is an infinitesimally small segment of the loop, \( \mu_0 \) is the permeability of free space, and \( I \) is the current.
This law is crucial for calculating magnetic fields in different configurations, including straight wires, coils, and solenoids. It simplifies into specific formulas for each case, making it a versatile tool in electromagnetism.
Circular Coil
A circular coil is composed of a wire wound in a circle, often with multiple turns, to create a uniform magnetic field inside the loop. When a current flows through the coil, each loop's magnetic field lines add up to form a stronger magnetic field inside.
For calculating the magnetic field at the center of a circular coil using Ampere's Law, we apply the formula:
  • \( B = \frac{\mu_0IN}{2r} \)
Where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space, \( I \) is the current, \( N \) is the number of turns, and \( r \) is the radius of the coil.
This setup is typically used in applications where a consistent and localized magnetic field is needed, such as in electromagnetic devices or sensors.
Solenoid
A solenoid is a long coil of wire, usually wrapped tightly in a helical form, with the primary purpose of generating a magnetic field along its axis. Solenoids are very effective at creating uniform magnetic fields inside, making them useful in many practical applications.
To find the magnetic field inside a solenoid, Ampere's Law is applied using the formula:
  • \( B = \frac{\mu_0IN}{L} \)
Here, \( B \) stands for the magnetic field, \( \mu_0 \) is the permeability of free space, \( I \) is the current, \( N \) is the number of turns, and \( L \) is the length of the solenoid.
Solenoids can be seen in devices like MRI machines and relays, where they contribute to the movement or the generation of strong, directed magnetic fields.
Magnetic Field of a Wire
The magnetic field of a straight wire carrying a current can be calculated using a specific form of Ampere's Law. For an infinitely long straight wire, the magnetic field forms concentric circles around the wire.
Ampere's Law provides us with the formula to compute this:
  • \( B = \frac{\mu_0I}{2\pi r} \)
Where \( B \) represents the magnetic field intensity at a distance \( r \) from the wire, \( \mu_0 \) is the permeability of free space, and \( I \) is the current flowing through the wire.
This principle is essential in understanding how current-carrying wires interact with their environment and how they can be used to generate magnetic fields in various applications, such as in inductive circuits and magnetic levitation systems.

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Most popular questions from this chapter

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 symmetric about the cylinder axis, is not constant but varies according to the relationship $$ \begin{array}{rlr} \overrightarrow{\boldsymbol{J}} & =\frac{2 I_{0}}{\pi a^{2}}\left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{k} & \text { for } r \leq a \\ & =\mathbf{0} \quad & \text { for } r \geq a \end{array} $$ where \(a\) is the radius of the cylinder, \(r\) is the radial distance from the cylinder axis, and \(I_{0}\) is a constant having units of amperes. (a) Show that \(I_{0}\) is the total current passing through the entire cross section of the wire. (b) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \geq a\). (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. (d) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \leq a\). How do your results in parts (b) and (d) compare for \(r=a ?\)

Two long, straight wires, one above the other, are separated by a distance \(2 a\) and are parallel to the \(x\) -axis. Let the \(+y\) -axis be in the plane of the wires in the direction from the lower wire to the upper wire. Each wire carries current \(I\) in the \(+x\) -direction. What are the magnitude and direction of the net magnetic field of the two wires at a point in the plane of the wires (a) midway between them; (b) at a distance \(a\) above the upper wire; (c) at a distance \(a\) below the lower wire?

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?

A long solenoid with 60 turns of wire per centimeter carries a current of 0.15 A. The wire that makes up the solenoid is wrapped around a solid core of silicon steel \(\left(K_{\mathrm{m}}=5200\right) .\) (The wire of the solenoid is jacketed with an insulator so that none of the current flows into the core.) (a) For a point inside the core, find the magnitudes of (i) the magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) due to the solenoid current; (ii) the magnetization \(\overrightarrow{\boldsymbol{M}}\) (iii) the total magnetic field \(\overrightarrow{\boldsymbol{B}}\). (b) In a sketch of the solenoid and core, show the directions of the vectors \(\overrightarrow{\boldsymbol{B}}, \overrightarrow{\boldsymbol{B}}_{0},\) and \(\overrightarrow{\boldsymbol{M}}\) inside the core.

The current in the windings of a toroidal solenoid is \(2.400 \mathrm{~A}\). There are 500 turns, and the mean radius is \(25.00 \mathrm{~cm} .\) The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be 1.940 T. Calculate (a) the relative permeability and (b) the magnetic susceptibility of the material that fills the toroid.
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