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A long, straight wire carries a current of 2.5 A. a) What is the strength of the magnetic field at a distance of \(3.9 \mathrm{~cm}\) from the wire? b) If the wire still carries \(2.5 \mathrm{~A}\), but is used to form a long solenoid with 32 turns per centimeter and a radius of \(3.9 \mathrm{~cm}\) what is the strength of the magnetic field at the center of the solenoid?

Short Answer

Expert verified
Question: Calculate the magnetic field at a distance of 3.9 cm from a straight wire carrying a current of 2.5 A. Then, calculate the magnetic field at the center of a solenoid with a given current of 2.5 A, 32 turns per centimeter, and a radius of 3.9 cm. Answer: The magnetic field at a distance of 3.9 cm from a straight wire carrying a current of 2.5 A is approximately \(2.56 \times 10^{-5} \mathrm{T}\). The magnetic field at the center of the solenoid is approximately \(0.0251 \mathrm{T}\).

Step by step solution

01

Part a: Magnetic field of a straight wire carrying current

The magnetic field (B) at a point near a straight wire carrying a current (I) can be found using the Ampere's law and Biot-Savart law, which gives us: B = \(\dfrac{\mu_0 I}{2\pi r}\), where B is the magnetic field strength, \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \mathrm{T.m/A}\)), I is the current flowing through the wire, and r is the distance from the wire. Given: I = \(2.5\mathrm{~A}\), r = \(3.9\mathrm{~cm}\) = \(0.039\mathrm{~m}\). Now, let's calculate the magnetic field strength (B) at the given distance (
02

Calculate B

): B = \(\dfrac{4\pi \times 10^{-7} \mathrm{T.m/A} \times 2.5\mathrm{~A}}{2\pi \times 0.039\mathrm{~m}}\) B = \(\dfrac{10^{-6} \mathrm{T.m^2}}{0.039\mathrm{~m}}\), B ≈ \(2.56 \times 10^{-5} \mathrm{T}\).
03

Part b: Magnetic field at the center of a solenoid

To find the magnetic field at the center of the solenoid, we can use the following formula: B = \(\mu_0 n I\), where B is the magnetic field strength at the center, n is the number of turns per unit length, I is the current through the solenoid, and \(\mu_0\) is the permeability of free space. Given: I = \(2.5 \mathrm{~A}\), n = 32 turns per centimeter = 3200 turns per meter, \(\mu_0\) = \(4\pi \times 10^{-7} \mathrm{T.m/A}\). Now, let's calculate the magnetic field strength at the center of the solenoid (
04

Calculate B

): B = \(4\pi \times 10^{-7} \mathrm{T.m/A} \times 3200 \mathrm{turns/m} \times 2.5 \mathrm{~A}\), B = \(0.008 \times 4\pi \mathrm{T}\), B ≈ \(0.0251 \mathrm{T}\). The magnetic field strength at the center of the solenoid is approximately \(0.0251\mathrm{T}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ampere's Law
When studying the behavior of magnetic fields, Ampere's Law stands out as one of the fundamental principles. It is a mathematical statement that relates the integrated magnetic field around a closed loop to the electric current passing through the loop.

Specifically, Ampere's Law can be expressed using the equation \( \oint B \cdot dl = \mu_0 I_{enc} \), where \( B \) is the magnetic field, \( dl \) is a differential element of the loop, \( \mu_0 \) denotes the permeability of free space, and \( I_{enc} \) is the current enclosed by the loop.

In the context of the problem provided, where a straight wire carries a current, Ampere's Law helps determine the magnetic field at some distance from the wire by simplifying into a more useful form. The law assumes a symmetrical setup, such as a long, straight wire, thereby allowing us to calculate the magnetic field easily by considering an imaginary loop around the wire.
Biot-Savart Law
While Ampere's Law provides an overall approach for calculating magnetic fields, the Biot-Savart Law offers a more detailed perspective, especially useful for points in space due to small segments of current-carrying wires. The law states that the magnetic field \( dB \) produced at a point by a small segment of current-carrying wire is related to the current \( I \) in the wire, the length of the segment \( dl \) and the position vector \( r \) from the segment to the point.

The formula \( dB = \frac{\mu_0}{4\pi} \frac{I dl \times \hat{r}}{r^2} \) captures this relationship, where \( \times \) represents the cross product, and \( \hat{r} \) is the unit vector in the direction of \( r \) – the distance vector from the current element to the point in question. For straight, long wires or loops like solenoids, this law lays the groundwork to arrive at expressions used for calculating the magnetic field strength.
Magnetic Field of Straight Wire
If you've ever held a compass near a wire through which current is flowing, you might've noticed the needle deflecting. This interaction is the result of the magnetic field generated by the current in the wire. The Biot-Savart Law can be used for deriving the magnetic field at any point around a long, straight wire, which is mathematically given by \( B = \frac{\mu_0 I}{2\pi r} \).

In this expression, \( B \) represents the magnetic field strength, \( \mu_0 \) the permeability of free space, \( I \) the current through the wire, and \( r \) the distance from the wire to the point where the field is measured. This straightforward equation is fundamental for solving problems related to magnetic fields around current-carrying wires, as demonstrated in the exercise we are discussing.
Magnetic Field at the Center of a Solenoid
A solenoid is essentially a coil of wire that is often wound in the shape of a helix. When an electric current flows through the wire, a uniform magnetic field is produced inside the solenoid, which is particularly strong along the central axis. This is a practical application where Ampere's Law becomes quite handy.

To calculate the magnetic field at the center of a solenoid, we apply the formula \( B = \mu_0 n I \), where \( n \) is the number of turns of the wire per unit length of the solenoid, and \( I \) is the current passing through it. The uniformity of the magnetic field inside a solenoid is a unique and useful characteristic that makes solenoids key components in many electromechanical devices. The exercise provided showcases how to compute this field using these very principles.

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Most popular questions from this chapter

You are standing at a spot where the magnetic field of the Earth is horizontal, points due northward, and has magnitude \(40.0 \mu \mathrm{T}\). Directly above your head, at a height of \(12.0 \mathrm{~m},\) a long, horizontal cable carries a steady \(\mathrm{DC}\) current of 500.0 A due northward. Calculate the angle \(\theta\) by which your magnetic compass needle is deflected from true magnetic north by the effect of the cable. Don't forget the sign of \(\theta-\) is the deflection eastward or westward?

An electron has a spin magnetic moment of magnitude \(\mu=9.285 \cdot 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\). Consequently, it has energy associated with its orientation in a magnetic field. If the difference between the energy of an electron that is "spin up" in a magnetic field of magnitude \(B\) and the energy of one that is "spin down" in the same magnetic field (where "up" and "down" refer to the direction of the magnetic field) is \(9.460 \cdot 10^{-25} \mathrm{~J}\), what is the field magnitude, \(B\) ?

A long, straight cylindrical tube of inner radius \(a\) and outer radius \(b\), carries a total current \(i\) uniformly across its cross section. Determine the magnitude of the magnetic field from the tube at the midpoint between the inner and outer radii.

The number of turns in a solenoid is doubled, and its length is halved. How does its magnetic field change? a) it doubles b) it is halved c) it quadruples d) it remains unchanged

Two long, straight wires are parallel to each other. The wires carry currents of different magnitudes. If the amount of current flowing in each wire is doubled, the magnitude of the force between the wires will be a) twice the magnitude of the original force. b) four times the magnitude of the original force. c) the same as the magnitude of the original force. d) half of the magnitude of the original force.

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