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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 40

(II) A wire is formed into the shape of two half circles connected by equal-length straight sections as shown in Fig. 28–48. A current I flows in the circuit clockwise as shown. Determine (a) the magnitude and direction of the magnetic field at the center, C, and (b) the magnetic dipole moment of the circuit.

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Step 1: Understand the geometry of the wire. The wire consists of two half-circles of radius R connected by two straight sections of equal length. The current flows clockwise through the circuit. The center of the wire, C, is at the center of the two half-circles.
Step 2: Calculate the magnetic field contribution at point C due to the two half-circles. Use the Biot-Savart law for a circular current loop. The magnetic field at the center of a full circular loop is given by: B = μI2R. Since each half-circle contributes half of this field, the magnetic field due to one half-circle is: Bhalf = μI4R. Add the contributions from both half-circles, keeping in mind their directions.
Step 3: Analyze the magnetic field contribution from the straight sections. The straight sections of the wire do not contribute to the magnetic field at point C because the current in these sections is parallel to the line connecting them to point C. The Biot-Savart law indicates that the magnetic field contribution from such sections at the center is zero.
Step 4: Determine the net magnetic field at point C. Combine the contributions from the two half-circles. Since the current flows clockwise, the magnetic field contributions from both half-circles at point C will add up in the same direction. Use the formula derived in Step 2 to calculate the total magnetic field: Btotal = 2 × μI4R = μI2R. The direction of the magnetic field can be determined using the right-hand rule.
Step 5: Calculate the magnetic dipole moment of the circuit. The magnetic dipole moment is given by: μ = IA, where A is the area enclosed by the circuit. The total area enclosed by the circuit is the sum of the areas of the two half-circles: A = 2 × πR22 = πR2. Substitute this into the formula for the magnetic dipole moment to find: μ = IπR2.

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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 Current

The magnetic field generated by an electric current is described by Ampère's Law, which states that the magnetic field around a closed loop is proportional to the current flowing through it. For a circular loop, the magnetic field at the center can be calculated using the formula B = (μ₀I)/(2R), where μ₀ is the permeability of free space, I is the current, and R is the radius of the loop.

Superposition of Magnetic Fields

When multiple current-carrying conductors are present, the total magnetic field at a point is the vector sum of the magnetic fields produced by each conductor. This principle of superposition allows us to analyze complex circuits by calculating the magnetic field contributions from each segment separately and then combining them to find the resultant field.

Magnetic Dipole Moment

The magnetic dipole moment (μ) is a vector quantity that represents the strength and orientation of a magnetic source. For a current loop, it is defined as μ = I * A, where I is the current and A is the area of the loop. The direction of the dipole moment is given by the right-hand rule, pointing perpendicular to the plane of the loop in the direction of the current.
Related Practice
Textbook Question

(III) Use the result of Problem 44 to find the magnetic field at point P in Fig. 28–53 due to the current in the square loop.


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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.


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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?


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Textbook Question

(III) A square loop of wire, of side d, carries a current I. (a) Determine the magnetic field B at points on a line (call it the 𝓍 axis) perpendicular to the plane of the square which passes through the center of the square (Fig. 28–56). Express B as a function of 𝓍, the distance from the center of the square. (b) For 𝓍 ≫ d, does the square appear to be a magnetic dipole? If so, what is its dipole moment?


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Textbook Question

(II) Consider a straight section of wire of length d, as in Fig. 28–51, which carries a current I. (a) Show that the magnetic field at a point P a distance 𝑅 from the wire along its perpendicular bisector is


B=μ0I2πRd(d2+4R2)12B = \(\frac{\mu_0 I}{2\pi R}\) \(\frac{d}{(d^2 + 4R^2)^{\frac{1}{2}\)}}


(b) Show that this is consistent with Example 28–10 for an infinite wire.

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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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