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Ch 28: Sources of Magnetic Field
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 28, Problem 40

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 55.0 cm long and 2.80 cm in diameter. What current will you need to produce the necessary field?

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Start by understanding the formula for the magnetic field inside a solenoid: B=μnI, where B is the magnetic field, μ is the permeability of free space (μ=4π×107Tm/A), n is the number of turns per unit length, and I is the current.
Calculate the number of turns per unit length, n, using the formula n=40000.55, where 4000 is the total number of turns and 0.55 m is the length of the solenoid.
Rearrange the formula to solve for the current I: I=B/μn. Substitute the values for B (0.150 T), μ (permeability of free space), and n (calculated in the previous step).
Ensure that all units are consistent, particularly converting the length of the solenoid from centimeters to meters, as the permeability constant is in meters.
After substituting the values into the rearranged formula, calculate the current I needed to produce the desired magnetic field.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Magnetic Field of a Solenoid

The magnetic field inside a solenoid is uniform and parallel to the axis of the solenoid. It is given by the formula B = μ₀ * n * I, where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. This formula is crucial for calculating the current needed to achieve a specific magnetic field strength.
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Permeability of Free Space

Permeability of free space, denoted as μ₀, is a fundamental physical constant that describes the ability of a vacuum to support a magnetic field. Its value is approximately 4π x 10⁻⁷ T·m/A. Understanding μ₀ is essential for calculating the magnetic field in a solenoid, as it directly influences the relationship between current and magnetic field strength.
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Number of Turns per Unit Length

The number of turns per unit length, n, is calculated by dividing the total number of turns by the length of the solenoid. It is a critical factor in determining the magnetic field inside the solenoid, as it directly affects the field's strength according to the formula B = μ₀ * n * I. For this problem, n = 4000 turns / 0.55 m.
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Related Practice
Textbook Question

A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It has radius 1.40 cm and length 40.0 cm, and the wire can carry a maximum current of 12.0 A. What total length of wire is required?

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Textbook Question

A closed curve encircles several conductors. The line integral Bdl\(\oint\) B\(\cdot\) dl around this curve is 3.83×104 T m3.83\(\times\)10^{-4}\(\text{ T m}\). If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.

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Textbook Question

A solid conductor with radius a is supported by insulating disks on the axis of a conducting tube with inner radius b and outer radius c (Fig. E28.43). The central conductor and tube carry equal currents I in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Derive an expression for the magnitude of the magnetic field at points outside the tube (r > c).

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Textbook Question

A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It has radius 1.40 cm and length 40.0 cm, and the wire can carry a maximum current of 12.0 A. What minimum number of turns per unit length must the solenoid have?

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Textbook Question

A 15.0 cm long solenoid with radius 0.750 cm is closely wound with 600 turns of wire. The current in the windings is 8.00 A. Compute the magnetic field at a point near the center of the solenoid.

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Textbook Question

Two long, parallel wires are separated by a distance of 0.400 m (Fig. E28.29). The currents I1 and I2 have the directions shown. Each current is doubled, so that I1 becomes 10.0 A and I2 becomes 4.00 A. Now what is the magnitude of the force that each wire exerts on a 1.20 m length of the other?

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