Problem 33 - University Physics with Modern Physics

Question

A closely wound, circular coil with a diameter of 4.00 cm has 600 turns and carries a current of 0.500 A. What is the magnitude of the magnetic field (a) at the center of the coil and (b) at a point on the axis of the coil 8.00 \(\mathrm{cm}\) from its center?

Step-by-Step Solution

Answer

The magnetic field at the center is approximately 9.42 mT, and at 8 cm from the center, it is approximately 66.2 μT.

Step 1: Calculate the Radius of the Coil

The diameter of the coil is given as 4.00 cm. To find the radius, divide the diameter by 2. \[ r = \frac{4.00}{2} = 2.00 \text{ cm} \] Since magnetic field calculations typically use SI units, we need to convert the radius to meters. Thus, 2.00 cm converts to:\[ r = 0.0200 \text{ m} \]

Step 2: Calculate the Magnetic Field at the Center of the Coil

The magnetic field at the center of a closely wound circular coil can be calculated using the formula:\[ B = \frac{\mu_0 \, n \, I}{2r} \] where:- \( B \) is the magnetic field,- \( \mu_0 = 4 \pi \times 10^{-7} \text{ Tm/A} \) is the permeability of free space,- \( n = 600 \) is the number of turns,- \( I = 0.500 \text{ A} \) is the current,- \( r = 0.0200 \text{ m} \) is the radius of the coil.Substitute the values into the formula:\[ B = \frac{(4 \pi \times 10^{-7}) \times 600 \times 0.500}{2 \times 0.0200} \]Calculate:\[ B = \frac{(4 \pi \times 10^{-7}) \times 300}{0.0400} \approx 9.42 \times 10^{-3} \text{ T} \]

Step 3: Calculate the Magnetic Field at a Point 8 cm from Coil Center

The magnetic field at a point along the axis of the coil at a distance \( x \) from the center can be calculated using:\[ B = \frac{\mu_0 \, n \, I \, r^2}{2(r^2 + x^2)^{3/2}} \]where:- \( x = 8.00 \text{ cm} = 0.0800 \text{ m} \) is the distance from the center.Substitute in the known values:\[ B = \frac{4 \pi \times 10^{-7} \times 600 \times 0.500 \times (0.0200)^2}{2((0.0200)^2 + (0.0800)^2)^{3/2}} \]Calculate:\[ B \approx 6.62 \times 10^{-5} \text{ T} \]

Key Concepts

Permeability of Free Space

The permeability of free space, denoted as \( \mu_0 \), is a fundamental physical constant that plays a crucial role in electromagnetism, particularly in the calculation of magnetic fields. It provides a measure of the ability of a vacuum to support magnetic fields.
\( \mu_0 \) is equal to \( 4\pi \times 10^{-7} \; \text{Tm/A} \), or Tesla meters per Ampere. This constant is used in the formula to calculate the magnetic field produced by a current-carrying conductor.
In our specific exercise, the permeability is key to finding the magnetic field strength both at the center of the coil and along its axis. Understanding \( \mu_0 \) is fundamental when dealing with any magnetic field calculations.

Magnetic Field at Center

To find the magnetic field at the center of a current-carrying coil, we use the formula:
\[ B = \frac{\mu_0 \, n \, I}{2r} \]
Here, \( B \) represents the magnetic field strength, \( n \) is the number of turns in the coil, \( I \) is the current passing through the coil, and \( r \) is the radius of the coil.
The position 'at the center' is significant as it implies the point where the magnetic field expression simplifies and the field is strongest for a circular loop.
In the exercise solution, substituting the given values like \( n = 600 \), \( I = 0.500 \) A, and \( r = 0.0200 \) m into the formula, we calculated \( B \approx 9.42 \times 10^{-3} \; \text{T} \). This result provides a clear example of how the formula operates with real-world values.

Magnetic Field on Coil Axis

Calculating the magnetic field on the axis of a coil is a bit more complex compared to the center. The field along the axis at a distance \( x \) from the center of the coil is given by:
\[ B = \frac{\mu_0 \, n \, I \, r^2}{2(r^2 + x^2)^{3/2}} \]
This formula accounts for the diminishing intensity of the magnetic field as you move away from the center of the coil.
Here, \( r \) is the radius of the coil, and \( x = 0.0800 \; \text{m} \) is the axial distance from the center.
Plugging the values into this formula in our solution, we found \( B \approx 6.62 \times 10^{-5} \; \text{T} \). This step is particularly useful for conceptualizing how magnetic fields behave not just at the central axis, but also at any point along the path of the coil axis.

Current-Carrying Coil

A current-carrying coil generates a magnetic field when electric current passes through it. The behavior of this magnetic field is crucial in many technological applications, ranging from electromagnets and inductors to transformers.
The strength and direction of the magnetic field depend on several factors like the number of turns, the current's magnitude, and the geometry of the coil.

• The more turns or loops the coil has, the stronger the magnetic field.
• The larger the current flowing through the coil, the greater the magnetic field produced.
• The size of the coil, specifically the radius, affects the calculation of the field strength.

These principles can be applied in calculating fields both at the center and at various points along the axis of the coil, as illustrated in our exercise. Understanding how a current-carrying coil works helps grasp broad concepts in electromagnetism and its numerous practical uses.