Question

A long solenoid (diameter of \(6.00 \mathrm{~cm}\) ) is wound with 1000 turns per meter of thin wire through which a current of 0.250 A is maintained. A wire carrying a current of 10.0 A is inserted along the axis of the solenoid. What is the magnitude of the magnetic field at a point \(1.00 \mathrm{~cm}\) from the axis?

Short answer

Answer: The magnitude of the combined magnetic field at a point 1.00 cm from the axis is 0.364 T.

Step-by-step solution

Find the magnetic field due to the solenoid

To find the magnetic field inside the solenoid, we will use the formula for the magnetic field of a solenoid: \(B_{solenoid} = \mu_0 * n * I\), where \(B_{solenoid}\) 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 flowing through the solenoid. Given, \(n = 1000 \, \text{turns/m}\), \(I = 0.250 \, \text{A}\). The permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} \,\text{Tm/A}\). Now, let's calculate \(B_{solenoid}\). \(B_{solenoid} = \mu_0 * n * I\)

Find the magnetic field due to the wire

The magnetic field generated by an infinitely long straight conductor at a distance \(r\) can be found by using the formula: \(B_{wire} = \frac{\mu_0 * I'}{2\pi r}\), where \(B_{wire}\) is the magnetic field, \(I'\) is the current in the wire, and \(r\) is the distance from the wire. Given, \(I' = 10.0 \, \text{A}\), and the distance \(r = 5.00 \, \text{cm} - 1.00 \,\text{cm} = 4.00 \,\text{cm} = 0.040 \, \text{m}\) (the radius of the solenoid minus the distance at which we want to calculate the magnetic field). Now, let's calculate \(B_{wire}\). \(B_{wire} = \frac{\mu_0 * I'}{2\pi r}\)

Calculate the combined magnetic field

Since both the solenoid and the wire have magnetic fields pointing in the same direction, we can add the magnitudes of these magnetic fields to get the combined magnetic field at the required point. \(B_{total} = B_{solenoid} + B_{wire}\) Now, substitute the values from Steps 1 and 2 to calculate the combined magnetic field.

Final calculation and answer

Now, we will substitute the given values and calculate the magnetic fields due to the solenoid and the wire, and then find the combined magnetic field. \(B_{solenoid} = (4\pi \times 10^{-7} \,\text{Tm/A}) * (1000 \, \text{turns/m}) * (0.250 \, \text{A}) = 0.314 \,\text{T}\) \(B_{wire} = \frac{(4\pi \times 10^{-7} \,\text{Tm/A}) * (10.0 \, \text{A})}{2\pi * 0.040 \, \text{m}} = 0.050 \,\text{T}\) Now, we can find the combined magnetic field: \(B_{total} = 0.314 \,\text{T} + 0.050 \,\text{T} = 0.364 \,\text{T}\) So, the magnitude of the magnetic field at a point \(1.00 \mathrm{~cm}\) from the axis is \(0.364 \,\text{T}\).

Key concepts

Solenoid Magnetic Field

A solenoid is essentially a coil of wire, often in cylindrical form, through which an electric current is passed to create a magnetic field. This magnetic field is particularly strong and uniform inside the solenoid itself. To simplify, think of it like a series of stacked loops, each contributing to the entire magnetic field along its axis.
When calculating the magnetic field inside the solenoid, we rely on the formula: \( B_{solenoid} = \mu_0 \times n \times I \). Here, \( \mu_0 \) represents the permeability of free space, \( n \) the number of turns per unit length, and \( I \) the current passing through the coil. The result demonstrates that the magnetic field is uniform and directly proportional to both the number of turns per unit length and the current.
This formula shows us that by adjusting the current or the density of turns, we can control the strength of the magnetic field within the solenoid.

• The formula highlights the importance of the solenoid's construction and its electrical current in determining the magnetic field.
• A larger current or denser turns lead to a stronger magnetic field.

Biot-Savart Law

The Biot-Savart Law provides a means to calculate the magnetic field produced by a current-carrying wire, helping us to understand how magnetic fields work in practical scenarios like that within solenoids or around wires. It states that the magnetic field \( B \) at a point in space is generated by an infinitely small piece of current-carrying wire. The law is key to finding how magnetic fields emanate from such wires.
For an infinitely long straight wire, the magnitude of the magnetic field felt at a distance \( r \) from the wire is given by the formula: \( B_{wire} = \frac{\mu_0 \times I'}{2\pi r} \). This tells us that the magnetic field strength decreases as you move further from the wire.
The Biot-Savart Law is instrumental in deriving the component due to the wire in the solenoid exercise, allowing us to account for the entirety of the interaction affecting magnetic fields in complex wire systems.

• It emphasizes the direct relationship between current and magnetic field strength.
• The field strength diminishes with increasing distance from the conductor.

Permeability of Free Space

The permeability of free space, noted as \( \mu_0 \), is a fundamental physical constant essential in electromagnetism, particularly in calculating the magnetic field in vacuum conditions. Its value is \( 4\pi \times 10^{-7} \, \text{Tm/A} \). This quantity explains how much resistance the vacuum offers to the formation of a magnetic field, hence influencing the magnetic environment.
In our solenoid magnetic field calculations, \( \mu_0 \) helps in determining how effectively the current and the solenoidal turns yield a magnetic field. This piece is crucial because it factors into both the magnetic field's strength within the solenoid (via the solenoid formula) and the field surrounding the wire (addressed through the Biot-Savart Law).

• \( \mu_0 \) is pivotal for the generation of magnetic fields without the interference of other materials.
• This constant serves as a fundamental underlying aspect across all electromagnetic calculations.