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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) 0.00904 T at solenoid center; (b) 3.5 x 10^-5 T for wire. Fields differ because the solenoid confines the field.

Step by step solution

01

Understand the Solenoid

A solenoid is a coil of wire designed to create a magnetic field when an electric current passes through it. The magnetic field inside a long solenoid is uniform and its strength is given by the formula: \( B = \mu_0 \frac{N}{L} I \), where \( B \) is the magnetic field, \( N \) is the number of turns, \( L \) is the length of the solenoid, \( I \) is the current, and \( \mu_0 = 4\pi \times 10^{-7} \, \mathrm{T \cdot m/A} \) is the permeability of free space.
02

Calculate the Magnetic Field Inside the Solenoid

Plug the given values into the formula for the magnetic field of the solenoid: \( N = 450 \), \( L = 0.35 \, \mathrm{m} \), \( I = 1.75 \, \mathrm{A} \). \[ B = \left( 4\pi \times 10^{-7} \right) \frac{450}{0.35} \times 1.75 \approx 0.00904 \, \mathrm{T} \] This is the magnetic field at the center of the solenoid.
03

Consider Magnetic Field 1 cm from Solenoid's Coil

Because the point is close to the solenoid's surface, the magnetic field is still approximately the same as calculated at the center within most of the solenoid, as the simple solenoid field formula does not account for distance from the coil in its interior where the formula holds.
04

Calculate the Magnetic Field of the Straight Wire

For a long straight wire carrying current, the magnetic field at a distance \( r \) is given by: \( B = \frac{\mu_0 I}{2\pi r} \). Given, \( r = 0.01 \, \mathrm{m} \) and \( I = 1.75 \, \mathrm{A} \), plug these values into the formula: \[ B = \frac{4\pi \times 10^{-7} \times 1.75}{2\pi \times 0.01} \approx 3.5 \times 10^{-5} \, \mathrm{T} \]
05

Compare the Magnetic Fields

The magnetic field calculated for the solenoid is approximately \( 0.00904 \, \mathrm{T} \), whereas the magnetic field for the straight wire is \( 3.5 \times 10^{-5} \, \mathrm{T} \). The magnetic fields are different. The solenoid confines the magnetic field inside it, and thus it is stronger and more uniform. The field produced by a straight wire decreases with distance.

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Key Concepts

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

Understanding Solenoids
A solenoid is essentially a long coil of wire, often cylindrical, where several turns are wrapped closely together. This setup is specifically designed to create a stronger and more uniform magnetic field. The key factor here is that when an electric current flows through the wire, it generates a magnetic field around it.

- **Uniform Magnetic Field:** Inside a solenoid, the magnetic field lines are parallel and evenly spaced. This uniformity is due to the closely packed circular loops of wire that contribute together to create a strong magnetic field in the interior of the solenoid. - **Practical Uses:** Solenoids are commonly used in devices such as electromagnets, starters in car engines, and valves in fluid systems.

A solenoid is crucial when there's a need for a compact and efficient magnetic field, as it increases the field strength compared to a straight wire carrying the same current.
Impact of Current in Wire
The current flowing through the wire in a solenoid or any other conductor is integral in determining the strength of the magnetic field generated. This occurs because, as per the laws of electromagnetism, an electric current induces a magnetic field.

- **Direct Proportionality:** The greater the current flowing through the solenoid's coil, the stronger the magnetic field produced. This is evident in the formula for magnetic field strength within a solenoid which includes current ( I ) as a direct multiplier. - **Consistency:** For both solenoids and wires, maintaining a consistent current is necessary for predictable and stable magnetic field results.

Current is a primary factor that must be controlled precisely in applications needing exact magnetic field strength, like MRI machines and certain types of sensors.
Magnetic Field Calculation Basics
When calculating the magnetic field strength, different approaches and formulas are used depending on the form and setup of the conductor. For a solenoid, the formula is:\[ B = \mu_0 \frac{N}{L} I \]Where \( B \) is the magnetic field, \( N \) is the number of turns, \( L \) is the length of the solenoid, \( I \) is the current, and \( \mu_0 \) is the permeability of free space.- **Parameters Affecting Field Strength:** - **Number of Turns (N):** More coils tightly packed usually result in a stronger magnetic field. - **Solenoid Length (L):** Longer solenoids tend to create a more uniform magnetic field. - **Current (I):** As highlighted earlier, increasing the current enhances the magnetic field. - **Permeability of Free Space (\( \mu_0 \)):** A constant value affecting how easily a magnetic field can form in a medium.
For a long straight wire, the magnetic field at a distance \( r \) from the wire is:\[ B = \frac{\mu_0 I}{2\pi r} \]Here, the magnetic field decreases as you move further from the wire, reflecting its less confined nature compared to a solenoid.
Ampère's Law and Its Application
Ampère's Law is a fundamental principle used to calculate magnetic fields in many situations, including solenoids and wires. It relates the magnetic field in a loop to the electric current passing through the loop.

- **Basic Formulation:** Ampère's Law states that the line integral of the magnetic field \( B \) around any closed loop is equal to the permeability of free space \( \mu_0 \) times the current \( I \) enclosed by that loop. Mathematically, it is expressed as:\[ \oint \mathbf{B} \cdot \mathrm{d}\mathbf{l} = \mu_0 I_{\text{enclosed}} \]- **Solenoids:** In the context of a solenoid, Ampère's Law helps explain why the magnetic field is concentrated and stronger inside the solenoid compared to outside.- **Straight Wires:** For a straight wire, this law helps in understanding how the magnetic field dissipates with distance from the wire.
Ampère's Law is invaluable in both theoretical studies and practical applications for determining magnetic field strength and understanding its behavior in complex systems.

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Most popular questions from this chapter

\bullet Bubble chamber, I. Certain types of bubble chambers are filled with liquid hydrogen. When a particle (such as an electron or a proton) passes through the liquid, it leaves a track of bubbles, which can be photographed to show the path of the particle. The apparatus is immersed in a known magnetic field, which causes the particle to curve. Figure 20.77 is a trace of a bubble chamber image showing the path of an electron. (a) How could you determine the sign of the charge of a particle from a photograph of its path? (b) How can physicists determine the momentum and the speed of this electron by using measurements made on the photograph, given that the magnetic field is known and is perpendicular to the plane of the figure? (c) The electron is obviously spiraling into smaller and smaller circles. What properties of the electron must be changing to cause this behavior? Why does this happen? (d) What would be the path of a neutron in a bubble chamber? Why?

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.

An electron moves at \(2.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) through a region in which there is a magnetic field of unspecified direction and magnitude \(7.40 \times 10^{-2} \mathrm{T}\) (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

\(\cdot\) The plane of a 5.0 \(\mathrm{cm}\) by 8.0 \(\mathrm{cm}\) rectangular loop of wire is parallel to a 0.19 T magnetic field, and the loop carries a cur- rent of 6.2 \(\mathrm{A}\) . (a) What torque acts on the loop? (b) What is the magnetic moment of the loop?

. You have 25 \(\mathrm{m}\) of wire, which you want to use to construct a 44 \(\mathrm{cm}\) diameter coil whose magnetic field at its center will exactly cancel the earth's field of 0.55 gauss. What current will your coil require?

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