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Chapter 20: Problem 66

A solenoid that is \(35 \mathrm{~cm}\) long and contains 450 circular coils \(2.0 \mathrm{~cm}\) in diameter carries a 1.75 A current. (a) What is the magnetic field at the center of the solenoid, \(1.0 \mathrm{~cm}\) from the coils? (b) Suppose we now stretch out the coils to make a very long wire carrying the same current as before. What is the magnetic field \(1.0 \mathrm{~cm}\) from the wire's center? Is it the same as you found in part (a)? Why or why not?

Short Answer

Expert verified
(a) 0.0028 T; (b) 0.000035 T; No, because solenoids concentrate magnetic fields effectively.

Step by step solution

01

Calculate the number of turns per unit length

Divide the total number of coils by the solenoid's length to find the number of turns per meter. The length of the solenoid is given as 35 cm (which is 0.35 m), and the number of turns is 450.The number of turns per meter \( n \) is:\[ n = \frac{450}{0.35} \approx 1285.71 \text{ turns/m} \]
02

Use the solenoid magnetic field formula

Use the formula for the magnetic field inside a solenoid:\[ B = \mu_0 n I \]where:- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \text{ T m/A} \)),- \( n \) is the number of turns per meter (from Step 1),- \( I \) is the current through the solenoid (1.75 A).Substitute the values:\[ B = (4\pi \times 10^{-7}) \times 1285.71 \times 1.75 \approx 0.0028 \text{ T} \]
03

Understand the scenario change for the wire

When the coils are stretched into a very long wire, it behaves like a long straight conductor. We need to find the magnetic field at 1.0 cm from this wire's center.
04

Use the formula for the magnetic field of a long straight conductor

The formula for the magnetic field at a distance \( r \) from a long straight wire carrying current \( I \) is:\[ B = \frac{\mu_0 I}{2\pi r} \]Here, \( r \) is 0.01 m (1.0 cm), and \( I \) is 1.75 A.Substitute the values:\[ B = \frac{4\pi \times 10^{-7} \times 1.75}{2\pi \times 0.01} = \frac{7\times 10^{-7}}{0.02} \approx 0.000035 \text{ T} \]
05

Compare the magnetic fields

Compare the results from step 2 and step 4. The magnetic field inside the solenoid (0.0028 T) is much stronger than the field around the straight wire (0.000035 T). This is due to the confined coil structure of a solenoid, which enhances the field strength.

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Key Concepts

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

Solenoid Physics
A solenoid is a type of electromagnet that is formed by coiling a wire into a cylindrical shape. When an electric current passes through the wire, it generates a magnetic field inside the coiled solenoid. The strength and stability of this field make solenoids a common choice in various devices, ranging from simple switches to complex machinery.

Key characteristics of solenoids include:
  • The magnetic field inside a solenoid is uniform and strong, especially when the coil is long and tightly wound.
  • The field lines are concentrated inside the coil, which limits the magnetic effects to the interior space.
  • Outside the solenoid, the magnetic field is weak and less orderly.
The interior magnetic field of a solenoid depends on the number of coils, the current flowing through the wire, and the permeability of the core material. A solenoid with more coils or higher current will produce a stronger magnetic field.

The solenoid's design makes it efficient for applications requiring a consistent magnetic field, such as in transformers, inductors, and electromagnets.
Magnetic Field Equations
The magnetic field inside a solenoid is given by the formula \( B = \mu_0 n I \), where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, and \( I \) is the current. This equation highlights how the magnetic field strength depends on these parameters.

Understanding each element:
  • \( \mu_0 \) is a constant that reflects the ability of the vacuum to support a magnetic field.
  • \( n \) represents how densely the wire is coiled; more turns per meter mean a more intense field.
  • \( I \), the current, shows direct correlation; more current leads to a stronger magnetic field.
Another critical equation deals with the magnetic field outside of a solenoid when the wire is straightened. The formula \( B = \frac{\mu_0 I}{2\pi r} \) describes the magnetic field at a distance \( r \) from a long straight conductor. This equation shows that the field's strength diminishes with distance and is inversely proportional to \( r \).

These equations are essential for engineers and scientists when designing and using solenoids in practical applications.
Long Straight Conductor Magnetic Field
When the coiled wire of a solenoid is straightened, it behaves like a long straight conductor. In this configuration, the distribution and strength of the magnetic field change significantly. Unlike the solenoid, which concentrates magnetic lines inside, a long straight conductor has its field distributed in circular loops around the wire.

Some characteristics of the magnetic field around a long straight conductor include:
  • The magnetic field forms concentric circles around the wire, with the field intensity decreasing as you move further away.
  • The strength of the magnetic field at any point is proportional to the current and inversely proportional to the distance from the wire.
  • This results in a weaker magnetic field at greater distances compared to the concentrated field of a solenoid.
Understanding these differences helps explain why the magnetic field near a long straight wire is different from that inside a solenoid. This concept is crucial, especially in power transmission and when analyzing electromagnetic effects in various electrical engineering applications.

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Most popular questions from this chapter

A circular metal loop is \(22 \mathrm{~cm}\) in diameter. (a) How large a current must flow through this metal so that the magnetic field at its center is equal to the earth's magnetic field of \(0.50 \times 10^{-4} \mathrm{~T} ?\) (b) Show how the loop should be oriented so that it can cancel the earth's magnetic field at its center.

The magnetic field around the human head has been measured to be approximately \(3.0 \times 10^{-8}\) gauss. Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modeling them as a single circular current loop \(16 \mathrm{~cm}\) (the width of a typical head) in diameter. What is the current needed to produce such a field at the center of the loop?

A straight vertical wire carries a current of 1.20 A downward in a region between the poles of a large electromagnet where the field strength is \(0.588 \mathrm{~T}\) and is horizontal. What are the magnitude and direction of the magnetic force on a \(1.00 \mathrm{~cm}\) section of this wire if the magnetic- field direction is (a) toward the east, (b) toward the south, (c) \(30.0^{\circ}\) south of west?

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near its center. You have enough wire for 4000 circular turns, and the solenoid must be \(1.40 \mathrm{~m}\) long and \(2.00 \mathrm{~cm}\) in diameter. What current will you need to produce the necessary field?

A particle with a charge of \(-2.50 \times 10^{-8} \mathrm{C}\) is moving with an instantaneous velocity of magnitude \(40.0 \mathrm{~km} / \mathrm{s}\) in the \(x\) - \(y\) plane at an angle of \(50^{\circ}\) counterclockwise from the \(+x\) axis. What are the magnitude and direction of the force exerted on this particle by a magnetic field with magnitude \(2.00 \mathrm{~T}\) in the (a) \(-x\) direction and (b) \(+z\) direction?
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