Problem 55 - Fundamentals of Physics

Question

A long solenoid with \(10.0\) turns \(/ \mathrm{cm}\) and a radius of \(7.00 \mathrm{~cm}\) carries a current of \(20.0 \mathrm{~mA}\). A current of \(6.00 \mathrm{~A}\) exists in a straight conductor located along the central axis of the solenoid. (a) At what radial distance from the axis will the direction of the resulting magnetic field be at \(45.0^{\circ}\) to the axial direction? (b) What is the magnitude of the magnetic field there?

Step-by-Step Solution

Answer

(a) Radial distance is approximately 0.048 m; (b) Magnitude is approximately 2.83 \( \mu \)T.

Step 1: Determine the magnetic field due to the solenoid

The magnetic field at any point inside a long solenoid is given by the formula \( B_s = \mu_0 n I_s \), where \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \ \text{Tm/A}) \), \( n \) is the number of turns per unit length, and \( I_s \) is the current through the solenoid. Given \( 10.0 \ \text{turns/cm} = 1000 \ \text{turns/m} \) and \( I_s = 20.0 \ \text{mA} = 0.020 \ \text{A} \), we can calculate \( B_s \).

Step 2: Calculate the magnetic field due to the straight conductor

The magnetic field due to a long straight conductor at a distance \( r \) is given by \( B_w = \frac{\mu_0 I_w}{2\pi r} \), where \( I_w = 6.00 \ \text{A} \) is the current in the wire. This will need to be combined later with the solenoid's field to find the position where the angle is \( 45^\circ \).

Step 3: Set up the condition for the angle

The problem states that the angle between the total magnetic field and the solenoid's axis is \( 45^\circ \). This implies that the tangential component \( B_w \) equals the axial component \( B_s \). Therefore, \( B_s = B_w \).

Step 4: Solve for the radial distance \( r \)

Substitute the expressions for \( B_s \) and \( B_w \) into \( B_s = B_w \): \( \mu_0 n I_s = \frac{\mu_0 I_w}{2\pi r} \). Cancel \( \mu_0 \) from both sides and solve for \( r \): \( r = \frac{I_w}{2\pi n I_s} \). Substitute \( I_w = 6.00 \ \text{A} \), \( n = 1000 \ \text{turns/m} \), and \( I_s = 0.020 \ \text{A} \) to find \( r \).

Step 5: Calculate the magnitude of the total magnetic field

Once \( r \) is known, substitute back into \( B_w = \frac{\mu_0 I_w}{2\pi r} \) to find the magnetic field \( B_w \) at that distance. Since \( B_s = B_w \), the magnitude of the total magnetic field is \( \sqrt{2} B_s \), combining \( B_w \) and \( B_s \) at \( 45^\circ \).

Key Concepts

Magnetic Field

The magnetic field is a fundamental aspect of electromagnetism. It is a vector field that exerts a force on moving charges, like electrons, and magnets. Magnetic fields are represented by field lines, where the density of the lines indicates the strength of the field. The direction of these lines is defined from the north to the south pole of a magnet.
In terms of interaction with electrical currents, the magnetic field surrounds any conductor carrying a current. This phenomenon is described by Ampère’s Law, which states that an electric current produces a magnetic field around it. The strength and direction of this field depend on the nature of the current and its path.
Mathematical expressions also apply: for example, the field due to a long, straight conductor is given by \( B = \frac{\mu_0 I}{2\pi r} \), where \( B \) is the magnetic field's magnitude, \( \mu_0 \) is the magnetic constant, \( I \) is the current, and \( r \) is the distance from the wire.

In our exercise, this concept is applied to both solenoids and straight conductors, showing how these fields interact and influence each other.

Solenoid

A solenoid is a coil of wire designed to generate a uniform magnetic field in its interior. The magnetic field inside a long solenoid that carries current is constant and parallel to the axis of the solenoid.
A solenoid’s magnetic field strength is determined using the formula: \( B_s = \mu_0 n I_s \), where \( n \) is the number of turns of wire per unit length, and \( I_s \) is the current running through the solenoid. The permeability of free space, \( \mu_0 \), is a constant that bridges the physical magnetic characteristics of free space with mathematical expressions.

In the context of this exercise, the solenoid creates an axial magnetic field. When paired with the magnetic field from the straight conductor, the resultant magnetic field is evaluated to determine a specific orientation. The solenoid’s ability to create such a predictable field makes it valuable in applications like electromagnets and inductors.

Straight Conductor

A straight conductor carrying current generates a circular magnetic field around it. According to the right-hand rule, if you point the thumb of your right hand in the direction of the current, your fingers will curl in the direction of the magnetic lines of force.
The strength of this field at a distance \( r \) from the wire is given by the formula \( B_w = \frac{\mu_0 I_w}{2\pi r} \), where \( I_w \) is the current through the wire. This formula shows that the magnetic field decreases with increasing distance from the wire and is directly proportional to the current.

Within the exercise, the straight conductor runs along the axes of the solenoid, and the resulting magnetic fields combine to produce a net effect that is analyzed at a specific radial position. This combination of fields from different sources demonstrates the interplay of simple magnetic structures and their investigation in more complex systems.