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A closed curve encircles several conductors. The line integral \(\oint \vec{B} \cdot d \vec{l}\) around this curve is \(3.83 \times 10^{-4} \mathrm{T} \cdot \mathrm{m}\) . (a) What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.

Short Answer

Expert verified
(a) The net current is approximately 304 A. (b) The line integral in the opposite direction is \(-3.83 \times 10^{-4} \ \mathrm{T} \cdot \mathrm{m}\).

Step by step solution

01

Understanding Ampere's Law

Ampere's Law states that the line integral of the magnetic field \( \vec{B} \) around a closed path is equal to the permeability of free space \( \mu_0 \) multiplied by the net current \( I_{enc} \) enclosed by the path. Mathematically, it is represented as: \( \oint \vec{B} \cdot d \vec{l} = \mu_0 I_{enc} \).
02

Identify Given Values

From the problem statement, we know that \( \oint \vec{B} \cdot d \vec{l} = 3.83 \times 10^{-4} \ \mathrm{T} \cdot \mathrm{m} \) and the permeability of free space is \( \mu_0 = 4\pi \times 10^{-7} \ \mathrm{T} \cdot \mathrm{m/A} \).
03

Solve for Net Current

Using Ampere's Law, divide both sides of the equation by \( \mu_0 \) to find the net current: \( I_{enc} = \frac{\oint \vec{B} \cdot d \vec{l}}{\mu_0} = \frac{3.83 \times 10^{-4}}{4\pi \times 10^{-7}} \).
04

Calculate Net Current

Perform the calculation: \( I_{enc} = \frac{3.83 \times 10^{-4}}{4\pi \times 10^{-7}} \approx 304 \ \mathrm{A} \). This is the net current enclosed by the curve.
05

Determine Effect of Opposite Direction

If the curve is integrated in the opposite direction, according to Ampere's Law, the line integral \( \oint \vec{B} \cdot d \vec{l} \) would have the same magnitude but an opposite sign. This means the integral becomes \( -3.83 \times 10^{-4} \ \mathrm{T} \cdot \mathrm{m} \).
06

Explanation

The negative sign occurs because the direction of integration is reversed, which reverses the direction of the integral vector \( d \vec{l} \). Thus, using Ampere's Law, the physical meaning remains consistent but changes sign due to the orientation.

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Key Concepts

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

magnetic field
Magnetic fields are invisible forces that act around magnetic materials and electric currents. Imagine this: whenever you see a magnet attracting things like paper clips, that's a magnetic field in action. Its core property is to exert a force on other magnets or current-carrying wires within its reach. This happens because electrons in the magnet are continuously moving, creating an unseen but impactful area of force.

- A magnetic field is represented by the symbol \( \vec{B} \).
- It has both a direction (shown by field lines) and a magnitude (how strong the field is).
- The strength is measured in Teslas (T).

Think of magnetic field lines like paths traced by iron filings around a magnet. They show the field's direction and density, where denser lines equal a stronger field. In physics problems, understanding these lines is key to predicting how magnetic forces will act on materials and currents nearby.
line integral
The line integral in the context of Ampere’s Law helps us find the total magnetic effect around a closed path. Simply put, it’s a way to measure how much the magnetic field \( \vec{B} \) "travels" along a given loop.

- Represented as \( \oint \vec{B} \cdot d \vec{l} \). This symbol means we are adding up all the tiny bits where the magnetic field lines follow a path.
- This integral is crucial in determining the influence of the magnetic field on that path.

When we compute this line integral, we consider how the magnetic field intersects with small segments of the path (\( d \vec{l} \)), adding these all up in small bits to form the total measurement. This approach is reminiscent of adding pieces of a puzzle together to complete a picture. If the path direction changes, the magnitude of the integral remains the same, but its sign flips because the direction of joining the dots has reversed.
net current
Net current is the total electric current flowing through all wires encircled by our path. Knowing this helps us make sense of how much electricity is in motion.

- The net current inside a loop is denoted as \( I_{enc} \).
- Its significance in Ampere's Law reveals how currents dictate the magnetic fields around them.

When you apply Ampere’s Law, the line integral of the magnetic field around the path gives you the enclosed net current once divided by the permeability of free space (\( \mu_0 \)). The Law is thus handy for calculating unknown currents if the magnetic field is known, making it an essential tool in electronics and electromagnetism:
  • A higher net current means a stronger magnetic effect.
  • This relationship allows us to understand the possible impact and design of circuit systems.
Understanding net current is key in electronic design to ensure systems are both efficient and safe, as it unravels the full picture of electromagnetism’s influence in circuits.

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Most popular questions from this chapter

Two long, parallel transmission lines, 40.0 \(\mathrm{cm}\) apart, carry \(25.0-\mathrm{A}\) and 75.0 -A currents. Find all locations where the net magnetic field of the two wires is zero if these currents are in (a) the same direction and (b) the opposite direction.

BIO EMF. Currents in de transmission lines can be 100 A or more. Some people have expressed concern that the electromagnetic fields (EMFs) from such lines near their homes could cause health dangers. For a line with current 150 \(\mathrm{A}\) and at a height of 8.0 \(\mathrm{m}\) above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percent of the earth's magnetic field, which is 0.50 gauss. Does this seem to be cause for worry?

A wooden ring whose mean diameter is 14.0 \(\mathrm{cm}\) is wound with a closely spaced toroidal winding of 600 turns. Compute the magnitude of the magnetic field at the center of the cross section of the windings when the current in the windings is 0.650 \(\mathrm{A}\) .

A square wire loop 10.0 \(\mathrm{cm}\) on each side carries a clockwise current of 15.0 A. Find the magnitude and direction of the magnetic field at its center due to the four 1.20 -mm wire segments at the midpoint of each side.

A long solenoid with 60 turns of wire per centimeter carries a current of 0.15 A. The wire that makes up the solenoid is wrapped around a solid core of silicon steel \(\left(K_{\mathrm{m}}=5200\right) .\) (The wire of the solenoid is jacketed with an insulator so that none of the current flows into the core.) (a) For a point inside the core, find the magnitudes of (i) the magnetic field \(\vec{B}_{0}\) due to the solenoid current; (ii) the magnetization \(\vec{M} ;\) (iii) the total magnetic field \(\vec{\boldsymbol{B}}\) . (b) In a sketch of the solenoid and core, show the directions of the vectors \(\vec{\boldsymbol{B}}, \vec{\boldsymbol{B}}_{0},\) and \(\vec{M}\) inside the core.

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