Ch. 28 - Sources of Magnetic Field

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 28 - Sources of Magnetic Field

Problem 21

Chapter 27, Problem 21

(II) Two long wires are oriented so that they are perpendicular to each other. At their closest, they are 20.0 cm apart (Fig. 28–42). What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 18.0 A and the bottom one carries 12.0 A?

Verified step by step guidance

Step 1: Understand the problem. Two long wires are perpendicular to each other, and we need to calculate the magnitude of the magnetic field at a point midway between them. The wires carry currents of 18.0 A and 12.0 A, respectively, and the closest distance between them is 20.0 cm.

Step 2: Recall the formula for the magnetic field due to a long straight current-carrying wire. The magnetic field at a distance $ r $ from a wire carrying current $ I $ is given by $ B = \frac{\mu_0 I}{2 \pi r} $, where $ \mu_0 $ is the permeability of free space ($ \mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A} $).

Step 3: Determine the distance from each wire to the point midway between them. Since the wires are perpendicular and the closest distance between them is 20.0 cm, the point midway between them forms a right triangle with the wires. The distance from each wire to the midpoint is $ r = \frac{20.0 \text{ cm}}{2} = 10.0 \text{ cm} = 0.10 \text{ m} $.

Step 4: Calculate the magnetic field produced by each wire at the midpoint using the formula $ B = \frac{\mu_0 I}{2 \pi r} $. Substitute $ I = 18.0 \, \text{A} $ and $ r = 0.10 \, \text{m} $ for the top wire, and $ I = 12.0 \, \text{A} $ and $ r = 0.10 \, \text{m} $ for the bottom wire.

Step 5: Combine the magnetic fields from both wires. Since the wires are perpendicular, their magnetic fields at the midpoint are also perpendicular. Use the Pythagorean theorem to find the resultant magnetic field: $ B_{\text{total}} = \sqrt{B_1^2 + B_2^2} $, where $ B_1 $ and $ B_2 $ are the magnetic fields due to the top and bottom wires, respectively.

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Key Concepts

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

Magnetic Field Due to a Current-Carrying Wire

A long straight wire carrying an electric current generates a magnetic field around it. The strength of this magnetic field (B) at a distance (r) from the wire is given by the formula B = (μ₀I)/(2πr), where μ₀ is the permeability of free space and I is the current. The direction of the magnetic field can be determined using the right-hand rule.

Superposition Principle

The superposition principle states that when multiple magnetic fields are present, the total magnetic field at a point is the vector sum of the individual fields. This means that to find the net magnetic field at a point, one must calculate the magnetic field contributions from each source and then add them together, taking into account their directions.

Vector Addition of Magnetic Fields

Magnetic fields are vector quantities, meaning they have both magnitude and direction. When calculating the resultant magnetic field from multiple sources, it is essential to consider both the magnitudes and the directions of the fields. This often involves breaking the fields into their components and using trigonometric functions to find the resultant vector.

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Related Practice

Textbook Question

(II) Two long parallel wires 8.20 cm apart carry 19.5-A dc currents in the same direction. Determine the magnetic field vector at a point P, 12.0 cm from one wire and 13.0 cm from the other. See Fig. 28–43. [Hint: Use the law of cosines. See Appendix A or inside rear cover.]

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Textbook Question

(II) An electron enters a uniform magnetic field B = 0.28 T at a 45° angle to $\vec{B}$ . Determine the radius r and pitch p (distance between loops) of the electron’s helical path assuming its speed is 2.2 x 10⁶ m/s. See Fig. 27–48.

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Textbook Question

"(II) A rectangular loop of wire is placed next to a straight wire, as shown in Fig. 28–40. There is a dc current of 3.5 A in both wires. Determine the magnitude and direction of the net force on the loop.

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Textbook Question

(II) Two long straight wires each carry a dc current I out of the page toward the viewer, Fig. 28–38. Indicate, with appropriate arrows, the direction of $\vec{B}$ at each of the points 1 to 6 in the plane of the page. State if the field is zero at any of the points.

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Textbook Question

(II) Let two long parallel wires, a distance d apart, carry equal dc currents I in the same direction. One wire is at 𝓍 = 0, the other at 𝓍 = d, Fig. 28–41. Determine $\vec{B}$ along the 𝓍 axis between the wires as a function of 𝓍.

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Textbook Question

(III) A coaxial cable consists of a solid inner conductor of radius R₁, surrounded by a concentric cylindrical tube of inner radius R₂ and outer radius R₃ (Fig. 28–45). The conductors carry equal and opposite currents I₀ distributed uniformly across their cross sections. Determine the magnetic field at a distance R from the axis for: (a) R < R₁; (b) R₁ < R < R₂; (c) R₂ < R < R₃; (d) R > R₃. (e) Let I₀ = 1.50 A, R₁ = 1.00 cm , R₂ = 2.00 cm , and R₃ = 2.50 cm Graph B from R = 0 to R = 3.00 cm.

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