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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

We can estimate the strength of the magnetic field of a refrigerator magnet in the following way: Imagine the magnet as a collection of current-loop magnetic dipoles. (a) Derive the force between two current loops with radius \(R\) and current \(I\) separated by distance \(d \ll R\). Very close to the wire its magnetic field is about the same as for an infinitely long wire, and Eq.( 28.11 ) can be used. (b) Using Eq. ( 28.17 ), express the current \(I\) in terms of the magnetic field at the middle of the loop, and express the radius \(R\) in terms of the area of the loop. In this way, derive an expression for the force \(F\) between two identical current loops separated by a small distance \(d\) in terms of their mutual area \(A\) and center magnetic field \(B\). (c) Rearrange your result to obtain an expression for the magnetic field of a dipole with area \(A\) in terms of the force \(F\) from an identical dipole separated by a small distance \(d\). (d) Now notice that the force it takes to separate one magnet from your refrigerator is nearly the same as the force it takes to separate two magnets stuck together. Estimate that force \(F\). (e) Estimate the area of a refrigerator magnet. (f) Assume that when these magnets are stuck together or to the refrigerator, they are separated by an effective distance \(d=25 \mu \mathrm{m}\). Use the formula derived above to estimate the magnetic field strength of the magnet.

A short current element \(d \vec{l}=(0.500 \mathrm{~mm}) \hat{\jmath}\) carries a current of 5.40 A in the same direction as \(d \vec{l}\). Point \(P\) is located at \(\vec{r}=(-0.730 \mathrm{~m}) \hat{\imath}+(0.390 \mathrm{~m}) \hat{\boldsymbol{k}}\). Use unit vectors to express the mag- netic field at \(P\) produced by this current element.

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?

Long, straight conductors with square cross sections and each carrying current \(I\) are laid side by side to form an infinite current sheet (Fig. \(\mathbf{P 2 8 . 6 9}\) ). The conductors lie in the \(x y\) -plane, are parallel to the \(y\) -axis, and carry current in the \(+y\) -direction. There are \(n\) conductors per unit length measured along the \(x\) -axis. (a) What are the magnitude and direction of the magnetic field a distance \(a\) below the current sheet? (b) What are the magnitude and direction of the magnetic field a distance \(a\) above the current sheet?

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