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What is the magnitude of the magnetic field inside a long, straight tungsten wire of circular cross section with diameter \(2.4 \mathrm{~mm}\) and carrying a current of \(3.5 \mathrm{~A}\), at a distance of \(0.60 \mathrm{~mm}\) from its central axis?

Short Answer

Expert verified
Answer: The magnitude of the magnetic field inside the tungsten wire at a distance of 0.60 mm from its central axis is approximately 2.92 x 10^-3 T.

Step by step solution

01

Write down the formula for the magnetic field inside a wire

To calculate the magnetic field at a distance r from the central axis of a long straight wire carrying a current I, we use Ampere's Law formula: \[B = \frac{\mu_{0} I r}{2 \pi R^2}\] where B is the magnetic field, \(\mu_{0}\) is the permeability of free space (\(4\pi \times 10^{-7} \mathrm{~T \cdot m / A}\)), I is the current, r is the distance from the central axis, and R is the radius of the wire. Keep in mind that the wire has a diameter of \(2.4 \mathrm{~mm}\), so the radius R is half of that: \[R = \frac{2.4 \mathrm{~mm}}{2} = 1.2 \mathrm{~mm} = 1.2 \times 10^{-3} \mathrm{~m}\]
02

Plug in the given values

Now, we can plug in all the given values into the equation and solve for B. The current I is \(3.5 \mathrm{~A}\), and the distance r from the central axis is \(0.60 \mathrm{~mm} = 0.60 \times 10^{-3} \mathrm{~m}\): \[B = \frac{4\pi \times 10^{-7} \mathrm{~T \cdot m / A} \cdot 3.5 \mathrm{~A} \cdot 0.60 \times 10^{-3} \mathrm{~m}}{2 \pi \cdot (1.2 \times 10^{-3} \mathrm{~m})^2}\]
03

Calculate the magnetic field magnitude

Now, simplify the expression and calculate the magnetic field magnitude: \[B = \frac{4\pi \times 10^{-7} \mathrm{~T \cdot m / A} \cdot 3.5 \mathrm{~A} \cdot 0.60 \times 10^{-3} \mathrm{~m}}{2 \pi \cdot 1.44 \times 10^{-6} \mathrm{~m^2}} = \frac{7 \times 10^{-7} \mathrm{~T \cdot m / A} \cdot 0.60 \times 10^{-3} \mathrm{~m}}{1.44 \times 10^{-6} \mathrm{~m^2}}\] \[B \approx 2.92 \times 10^{-3} \mathrm{~T}\] So, the magnitude of the magnetic field inside the tungsten wire at a distance of \(0.60 \mathrm{~mm}\) from its central axis is approximately \(2.92 \times 10^{-3} \mathrm{~T}\).

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Key Concepts

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

Ampere's Law
Ampere's Law is a fundamental principle used to relate the magnetic field around a closed loop to the current passing through it. Essentially, Ampere’s Law states that the magnetic field integrated along a closed path is equal to the permeability of free space times the current passing through the path. This law is mathematically expressed as:
  • \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I \)
Here, \( \mathbf{B} \) is the magnetic field, \( d\mathbf{l} \) is an infinitesimal element of the path, \( \mu_0 \) is the permeability of free space, and \( I \) is the current enclosed by the loop.
In situations involving long, straight conductors, Ampere's Law simplifies the calculation of magnetic fields, especially for symmetrical configurations like cylindrical wires. By choosing an Amperian loop that fits the symmetry of the wire, the magnetic field can be calculated accurately.
Current in Wire
The current flowing through a conductor, such as a wire, is the rate at which electric charge passes through a cross-sectional area of the wire. It is usually denoted by \( I \) and measured in amperes (A). In this problem, we're dealing with a straight tungsten wire carrying a current of \( 3.5 \) A.
This current generates a magnetic field around the wire due to the movement of charges. Current in a wire can be visualized as a stream of electrons moving through the conductor, and the amount of current depends on two main factors:
  • The number of charges passing through the wire per unit time.
  • The charge each of these particles carries.
Understanding the relationship between current and magnetism is key to calculating the magnetic field within or around the wire.
Permeability of Free Space
Permeability of free space, symbolized as \( \mu_0 \), is a constant that quantifies the ability of a vacuum to sustain a magnetic field. The value of \( \mu_0 \) is approximately \( 4\pi \times 10^{-7} \) T·m/A.
This constant is vital in calculating magnetic fields using Ampere's Law. It reflects how much magnetic field is produced in a vacuum by a given current, serving as a bridge between electric currents and the magnetic fields they generate.
In essence, permeability of free space helps understand how electromagnetic fields propagate in a vacuum, playing a crucial role in many electromagnetic calculations.
Magnetic Field Calculation
To calculate the magnetic field inside a wire, we use the formula derived from Ampere's Law:
  • \( B = \frac{\mu_{0} I r}{2 \pi R^2} \)
Here, \( B \) is the magnetic field, \( I \) is the current, \( r \) is the distance from the central axis, and \( R \) is the wire's radius stored as \( 1.2 \times 10^{-3} \) m for this particular problem.
This formula enables the understanding of how the magnetic field varies within the wire. The magnetic field is directly proportional to the current in the wire and the radial distance \( r \), yet inversely proportional to the square of the radius \( R \).
By substituting known values into this equation, you can find the magnetic field's strength at any point inside the wire, as demonstrated in the given problem. This step-by-step approach ensures clarity in calculation.

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Most popular questions from this chapter

The magnetic dipole moment of the Earth is approximately \(8.0 \cdot 10^{22} \mathrm{~A} \mathrm{~m}^{2}\). The source of the Earth's magnetic field is not known; one possibility might be the circulation culating ions move a circular loop of radius \(2500 \mathrm{~km}\). What “current" must they produce to yield the observed field?

A loop of wire of radius \(R=25.0 \mathrm{~cm}\) has a smaller loop of radius \(r=0.900 \mathrm{~cm}\) at its center such that the planes of the two loops are perpendicular to each other. When a current of \(14.0 \mathrm{~A}\) is passed through both loops, the smaller loop experiences a torque due to the magnetic field produced by the larger loop. Determine this torque assuming that the smaller loop is sufficiently small so that the magnetic field due to the larger loop is same across the entire surface.

A hairpin configuration is formed of two semiinfinite straight wires that are \(2.00 \mathrm{~cm}\) apart and joined by a semicircular piece of wire (whose radius must be \(1.00 \mathrm{~cm}\) and whose center is at the origin of \(x y z\) -coordinates). The top straight wire is along the line \(y=1.00 \mathrm{~cm},\) and the bottom straight wire is along the line \(y=-1.00 \mathrm{~cm} ;\) these two wires are in the left side \((x<0)\) of the \(x y\) -plane. The current in the hairpin is \(3.00 \mathrm{~A},\) and it is directed toward the right in the top wire, clockwise around the semicircle, and to the left in the bottom wire. Find the magnetic field at the origin of the coordinate system.

Suppose that the magnetic field of the Earth were due to a single current moving in a circle of radius \(2.00 \cdot 10^{3} \mathrm{~km}\) through the Earth's molten core. The strength of the Earth's magnetic field on the surface near a magnetic pole is about \(6.00 \cdot 10^{-5} \mathrm{~T}\). About how large a current would be required to produce such a field?

Two particles, each with charge \(q\) and mass \(m\), are traveling in a vacuum on parallel trajectories a distance \(d\) apart, both at speed \(v\) (much less than the speed of light). Calculate the ratio of the magnitude of the magnetic force that each exerts on the other to the magnitude of the electric force that each exerts on the other: \(F_{\mathrm{m}} / \mathrm{F}_{\mathrm{e}}\)

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