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Ch 28: Sources of Magnetic Field
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 28, Problem 41b

A closed curve encircles several conductors. The line integral Bdl\(\oint\) B\(\cdot\) dl around this curve is 3.83×104 T m3.83\(\times\)10^{-4}\(\text{ T m}\). If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.

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1
Understand the concept of line integral of a magnetic field: The line integral of the magnetic field \( \oint \mathbf{B} \cdot d\mathbf{l} \) around a closed curve is related to the net current enclosed by the curve, according to Ampère's Law.
Recognize the significance of direction: The direction of integration around the curve affects the sign of the line integral. If you reverse the direction of integration, the sign of the integral will also reverse.
Apply Ampère's Law: Ampère's Law states that \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc} \), where \( I_{enc} \) is the net current enclosed by the curve. The given integral value is \( 3.83 \times 10^{-4} \) T m.
Consider the effect of reversing the direction: If the original integral value is positive, reversing the direction of integration will make the integral value negative, and vice versa.
Conclude the result: Therefore, if you integrate around the curve in the opposite direction, the value of the line integral will be \( -3.83 \times 10^{-4} \) T m.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ampère's Law

Ampère's Law relates the magnetic field around a closed loop to the electric current passing through the loop. It states that the line integral of the magnetic field B along a closed path is equal to the permeability of free space times the total current enclosed by the path. This principle is crucial for understanding how the magnetic field behaves around conductors.
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Line Integral

A line integral involves integrating a vector field along a curve. In the context of magnetic fields, it calculates the total magnetic influence along a path. The direction of integration affects the sign of the result, as reversing the path direction changes the sign of the integral, which is essential for understanding the problem's requirement to integrate in the opposite direction.
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Direction of Integration

The direction of integration in a line integral is significant because it determines the sign of the result. When integrating a vector field like the magnetic field, reversing the direction of the path changes the sign of the integral. Thus, integrating in the opposite direction would yield the negative of the original value, which is a key aspect of the question.
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Related Practice
Textbook Question

Four long, parallel power lines each carry 100 A currents. A cross-sectional diagram of these lines is a square, 20.0 cm on each side. For each of the three cases shown in Fig. E28.25, calculate the magnetic field at the center of the square.

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Textbook Question

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 55.0 cm long and 2.80 cm in diameter. What current will you need to produce the necessary field?

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Textbook Question

A solid conductor with radius a is supported by insulating disks on the axis of a conducting tube with inner radius b and outer radius c (Fig. E28.43). The central conductor and tube carry equal currents I in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Derive an expression for the magnitude of the magnetic field at points outside the tube (r > c).

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Textbook Question

A 15.0 cm long solenoid with radius 0.750 cm is closely wound with 600 turns of wire. The current in the windings is 8.00 A. Compute the magnetic field at a point near the center of the solenoid.

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Textbook Question

Two long, parallel wires are separated by a distance of 0.400 m (Fig. E28.29). The currents I1 and I2 have the directions shown. Each current is doubled, so that I1 becomes 10.0 A and I2 becomes 4.00 A. Now what is the magnitude of the force that each wire exerts on a 1.20 m length of the other?

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Textbook Question

Two long, parallel wires are separated by a distance of 0.400 m (Fig. E28.29). The currents I1 and I2 have the directions shown. Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other. Is the force attractive or repulsive?

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