Problem 68 - College Physics

Question

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?

Step-by-Step Solution

Answer

(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 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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: \[ B = \frac{4\pi \times 10^{-7} \times 1.75}{2\pi \times 0.01} \approx 3.5 \times 10^{-5} \, \mathrm{T} \]

Step 5: 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.

Key Concepts

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.