Explanation 1) Concept: The net magnetic field at the center of the square is the vector sum of the magnetic fields of all 4 wires. We can determine the magnetic field of each wire by using the formula for magnetic field at some point outside the current carrying wire. Adding them will give the magnetic field at the center of the square. 2) Formula: B = μ 0 i 2 π R 3) Given: i) The current carrying wires are perpendicular to the page ii) The current in each wire = 20 A iii) The distance between wires = edge length of square = 20 cm = 0.2 m iv) The currents in wires 1 and 4 are out of page, and those in wires 2 and 3 are into the page. 4) Calculations: a) The magnetic field at the center of the square is the vector sum of the magnetic field of all 4 wires. We use the Biot-Savart law to determine the magnetic field of the wires. For all the wires, the distance of the center from the wire is the same = R. It is related to edge length as R = d i a g o n a l 2 = a 2 2 = a 2 The magnetic field due to wire 1 is calculated as B = μ 0 i 2 π R ∴ B 1 = μ 0 i 2 π ( a 2 ) = μ 0 i 2 2 π a The current in the wire 1 is directed out of the page. Hence the direction of the magnetic field at the center is determined using the right hand rule. According to the rule, the direction of this field will be in the counter clockwise direction. So at the center it will be along the diagonal of the square, perpendicular to R. Hence it will have components along +x and +Y direction. The magnitude of the magnetic field will be the same for wire 4. The direction of this field will be in the counter clockwise direction. But at the center, it will be along the diagonal such that it has components along -X and +Y direction (as shown in the diagram). The magnitude of the magnetic field will be the same for wire 2 and wire 3 as well. But for these wires, the current is going into the page. The direction of this field will be in the clockwise direction. At the center, the field due to wire 2 will be same as B 4 and that for wire 3 will be same as B 1 . Looking at the diagram, we can see that the x components of the four fields will be directed in opposite directions along the X axis. Hence it will get cancelled. The Y components of all the fields are along the same direction (+Y). Hence resultant field will be the sum of these components

11th Edition

ISBN: 9781119472780

Author: Halliday

Publisher: WILEY

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Magnetic Force

A magnetic force arises when a charged particle or a current-carrying conductor is placed in a magnetic field. It is created because of the movement of the charged particles.

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Chapter 29, Problem 29P

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The magnetic field at the center of the square

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Chapter 29, Problem 28P

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Chapter 29 Solutions

FUNDAMENTALS OF PHYSICS,AP ED.

Ch. 29 - Prob. 1Q Ch. 29 - Prob. 2Q Ch. 29 - Prob. 3Q Ch. 29 - Prob. 4Q Ch. 29 - Prob. 5Q Ch. 29 - Prob. 6Q Ch. 29 - Prob. 7Q Ch. 29 - Prob. 8Q Ch. 29 - Prob. 9Q Ch. 29 - Prob. 10Q

Ch. 29 - Prob. 11Q Ch. 29 - A surveyor is using a magnetic compass 6.1 m below...Ch. 29 - Figure 29-35a shows an element of length ds = 1.00...Ch. 29 - SSM At a certain location in the Philippines,...Ch. 29 - Prob. 4P Ch. 29 - Prob. 5P Ch. 29 - Prob. 6P Ch. 29 - Prob. 7P Ch. 29 - Prob. 8P Ch. 29 - Prob. 9P Ch. 29 - Prob. 10P Ch. 29 - Prob. 11P Ch. 29 - Prob. 12P Ch. 29 - Prob. 13P Ch. 29 - Equation 29-4 gives the magnitude B of the...Ch. 29 - Prob. 15P Ch. 29 - Prob. 16P Ch. 29 - Prob. 17P Ch. 29 - Prob. 18P Ch. 29 - Prob. 19P Ch. 29 - Prob. 20P Ch. 29 - Prob. 21P Ch. 29 - Prob. 22P Ch. 29 - Prob. 23P Ch. 29 - Prob. 24P Ch. 29 - Prob. 25P Ch. 29 - Prob. 26P Ch. 29 - Prob. 27P Ch. 29 - GO Figure 29.56a shows two wires, each carrying....Ch. 29 - Prob. 29P Ch. 29 - Prob. 30P Ch. 29 - Prob. 31P Ch. 29 - GO The current-carrying wire loop in Fig. 29-6a...Ch. 29 - Prob. 33P Ch. 29 - Prob. 34P Ch. 29 - Prob. 35P Ch. 29 - Prob. 36P Ch. 29 - Prob. 37P Ch. 29 - Prob. 38P Ch. 29 - Prob. 39P Ch. 29 - Prob. 40P Ch. 29 - Prob. 41P Ch. 29 - Prob. 42P Ch. 29 - Prob. 43P Ch. 29 - Prob. 44P Ch. 29 - Prob. 45P Ch. 29 - Prob. 46P Ch. 29 - ILW The current density inside a long, solid,...Ch. 29 - Prob. 48P Ch. 29 - Prob. 49P Ch. 29 - A solenoid that is 95.0 cm long has a radius of...Ch. 29 - A 200-turn solenoid having a length of 25 cm and a...Ch. 29 - A solenoid 1.30 m long and 2.60 cm in diameter...Ch. 29 - A long solenoid has 100 turns/cm and carries...Ch. 29 - An electron is shot into one end of a solenoid. As...Ch. 29 - Prob. 55P Ch. 29 - Prob. 56P Ch. 29 - Prob. 57P Ch. 29 - Prob. 58P Ch. 29 - Prob. 59P Ch. 29 - Prob. 60P Ch. 29 - A circular loop of radius 12 cm carries a current...Ch. 29 - Prob. 62P Ch. 29 - Prob. 63P Ch. 29 - Prob. 64P Ch. 29 - A cylindrical cable of radius 8.00 mm carries a...Ch. 29 - Two long wires lie in an xy plane, and each...Ch. 29 - Two wires, both of length L, are formed into a...Ch. 29 - Prob. 68P Ch. 29 - Prob. 69P Ch. 29 - Prob. 70P Ch. 29 - A 10-gauge bare copper wire 2.6 mm in diameter can...Ch. 29 - A long vertical wire carries an unknown current....Ch. 29 - Prob. 73P Ch. 29 - The magnitude of the magnetic field at a point...Ch. 29 - Prob. 75P Ch. 29 - Prob. 76P Ch. 29 - Prob. 77P Ch. 29 - A long wire carrying 100 A is perpendicular to the...Ch. 29 - A long, hollow, cylindrical conductor with inner...Ch. 29 - A long wire is known to have a radius greater than...Ch. 29 - Prob. 81P Ch. 29 - Prob. 82P Ch. 29 - Prob. 83P Ch. 29 - Three long wires all lie in an xy plane parallel...Ch. 29 - Prob. 85P Ch. 29 - Prob. 86P Ch. 29 - Prob. 87P Ch. 29 - Prob. 88P

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The magnetic field at the center of the square

Solution Summary: The author calculates the magnetic field at the center of the square using the Biot-Savart law.

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To find: The magnetic field at the center of the square

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Chapter 29 Solutions

To find:

The magnetic field at the center of the square