Skip to main content
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
Chapter 27, Problem 23

(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.]

Verified step by step guidance
1
Step 1: Understand the problem. Two parallel wires carry currents in the same direction, and we need to find the net magnetic field at a point P located at specific distances from each wire. The magnetic field due to a current-carrying wire is given by the Biot-Savart law, simplified for a long straight wire as: B = μI2πr, where μ is the permeability of free space, I is the current, and r is the distance from the wire.
Step 2: Calculate the magnetic field contributions from each wire at point P. For the first wire, use the formula B1 = μI2πr1, where r1 is the distance from the first wire (12.0 cm). Similarly, for the second wire, use B2 = μI2πr2, where r2 is the distance from the second wire (13.0 cm).
Step 3: Determine the direction of the magnetic fields. Use the right-hand rule: point your thumb in the direction of the current, and your fingers curl in the direction of the magnetic field. Since the currents are in the same direction, the magnetic fields at point P will have different orientations. Represent these fields as vectors and note their directions.
Step 4: Use vector addition to find the net magnetic field at point P. The angle between the two magnetic field vectors can be determined using the law of cosines. The distance between the two wires is 8.20 cm, so the angle θ can be calculated as: cosθ = r12 + r22 - d22r1r2, where d is the distance between the wires (8.20 cm).
Step 5: Combine the magnetic field vectors. Use the formula for the resultant of two vectors: Bnet = B12 + B22 + 2B1B2cosθ. This will give the magnitude of the net magnetic field at point P. The direction can be found using trigonometry if needed.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
9m
Was this helpful?

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

The magnetic field generated by a long straight wire carrying a current can be described by Ampère's Law. The magnetic field (B) at a distance (r) from a long straight 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 follows the right-hand rule, curling around the wire in a circular pattern.
Recommended video:
Guided course
09:57
Magnetic Force on Current-Carrying Wire

Superposition of Magnetic Fields

When multiple current-carrying wires are present, the total magnetic field at a point is the vector sum of the magnetic fields produced by each wire. This principle of superposition allows us to calculate the resultant magnetic field by considering both the magnitude and direction of the fields from each wire at the point of interest.
Recommended video:
Guided course
05:30
Magnetic Fields and Magnetic Dipoles

Law of Cosines

The Law of Cosines is a mathematical formula used to relate the lengths of the sides of a triangle to the cosine of one of its angles. It is particularly useful in physics for determining distances and angles in problems involving vectors. In this context, it helps to find the resultant distance from the point of interest to the wires, which is essential for calculating the magnetic field.
Recommended video:
Related Practice
Textbook Question

(II) An electron enters a uniform magnetic field B = 0.28 T at a 45° angle to B\(\overrightarrow{B}\). 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.


1226
views
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.


1101
views
Textbook Question

(II) A circular conducting ring of radius 𝑅 is connected to two exterior straight wires at two ends of a diameter (Fig. 28–47). The current I splits into unequal portions as shown (unequal resistance) while passing through the ring. What is B\(\overrightarrow{B}\) at the center of the ring?


862
views
Textbook Question

(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?

510
views
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 B\(\overrightarrow{B}\) along the 𝓍 axis between the wires as a function of 𝓍.

830
views
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.

994
views