Problem 32 - University Physics with Modern Physics

Question

A closely wound, circular coil with radius 2.40 \(\mathrm{cm}\) has 800 turns. a) What must the current in the coil be if the magnetic field at the center of the coil is 0.0580 \(\mathrm{T}\) ? b) At what distance \(x\) from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?

Step-by-Step Solution

Answer

a) The current in the coil must be approximately 27.7 A. b) The distance from the center where the field is half is about 4.2 cm.

Step 1: Understanding the Formula for Magnetic Field of a Coil

The magnetic field at the center of a closely wound circular coil is given by the formula \( B = \frac{\mu_0 n I}{2R} \), where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space \((4\pi \times 10^{-7} \ \text{T}\cdot\text{m/A})\), \( n \) is the number of turns, \( I \) is the current, and \( R \) is the radius of the coil.

Step 2: Solving for Current I

We need to rearrange the equation for \( I \). So, \( I = \frac{2BR}{\mu_0 n} \). Substituting the given values: \( B = 0.0580 \ \text{T} \), \( R = 0.0240 \ \text{m} \), and \( n = 800 \), we get: \( I = \frac{2 \times 0.0580 \times 0.0240}{4\pi \times 10^{-7} \times 800} \). Calculate this to find the current.

Step 3: Calculating Current I

Substitute the values into the rearranged formula: \( I = \frac{2 \times 0.0580 \times 0.0240}{4\pi \times 10^{-7} \times 800} = \frac{0.002784}{1.0053 \times 10^{-4}} \approx 27.7 \ \text{A} \). So, the current needed is approximately 27.7 A.

Step 4: Using the Formula for the Magnetic Field on the Axis of a Coil

The formula for the magnetic field at a point on the axis of the coil is given by \(B_x = \frac{\mu_0 n I R^2}{2(R^2 + x^2)^{3/2}}\). We need this field \(B_x\) to be half the field at the center, i.e., \(B_x = \frac{B}{2} = \frac{0.0580}{2} = 0.0290 \ \text{T}\).

Step 5: Rearranging to Solve for Distance x

Set the equation \(0.0290 = \frac{4\pi \times 10^{-7} \times 800 \times 27.7 \times (0.024)^2}{2((0.024)^2 + x^2)^{3/2}}\) and solve for \(x\). Rearranging gives \((0.024)^2 + x^2 = \left(\frac{4\pi \times 10^{-7} \times 800 \times 27.7 \times (0.024)^2}{2 \times 0.0290}\right)^{2/3}\).

Step 6: Solving for Distance x

Calculate \( x \) using the rearranged equation. First calculate the right side, then subtract \((0.024)^2\) and finally take the square root to solve for \( x \). Completing the calculations, find that \( x \approx 0.042 \ \text{m} \) or 4.2 cm.

Key Concepts

Circular Coil

A circular coil is simply a loop of wire in a circular shape, which aids in creating a magnetic field when an electric current passes through it. These are widely used in different electromagnetic applications, such as inductors, transformers, and magnetic field sensing devices.

The coil in this exercise is described as "closely wound" with "800 turns." This means that the coil has 800 loops of wire, all packed tightly together, enhancing its ability to produce a strong magnetic field.

The radius of our circular coil is given as 2.40 cm. When considering coils, the radius plays a crucial role in determining how strong the magnetic field will be at specific points, like the center of the coil or on its axis.

Current in Coil

In electromagnetism, the current flowing through a coil significantly affects the magnetic field produced by the coil. The amount of current, in Amperes (A), determines the strength and shape of the magnetic field.

In our problem, the task is to find the required current for a magnetic field of 0.0580 T at the center of the coil. Using the formula: \[ B = \frac{\mu_0 n I}{2R} \]we can solve for the current. Rearranging gives: \[ I = \frac{2BR}{\mu_0 n} \]By substituting the known values, such as the magnetic field, the coil's radius, number of turns, and the permeability of free space, we find that a current of approximately 27.7 A is necessary.

Magnetic Field on Axis

The magnetic field on the axis of a coil differs from the field at the center. In the context of the exercise, it refers to the magnetic field strength at a point along the line that runs perpendicular to the plane of the coil and goes through the center of the coil.

The formula for the magnetic field at a point on the axis is:\[ B_x = \frac{\mu_0 n I R^2}{2(R^2 + x^2)^{3/2}} \]This formula tells us how the field strength decreases as we move away from the coil's center along its axis. The problem asks us to find how far we must go along the axis for the field to be half its center value. Solving this involves balancing the reduced field on the axis with specific values calculated.

Magnetic Field at Center

The magnetic field at the center of the coil is typically the strongest. This is because the magnetic field lines converge most tightly there. For a coil with multiple turns, like our 800-turn example, the field is enhanced by the cumulative effect of all the wire loops.

Mathematically, the center magnetic field is derived using the formula mentioned earlier, \[ B = \frac{\mu_0 n I}{2R} \]This relation helps in determining the field if you know the coil's current, number of turns, and its radius. As derived, the problem shows that a specific current through the coil is required to achieve the magnetic field strength desired at the center. The dependence on these parameters highlights the coil's efficiency in generating fields.