Chapter 22, Problem 70PE

### College Physics

1st Edition
Paul Peter Urone + 1 other
ISBN: 9781938168000

Chapter
Section

### College Physics

1st Edition
Paul Peter Urone + 1 other
ISBN: 9781938168000
Textbook Problem

# Find the magnitude and direction of the magnetic field at the point equidistant from the wires in Figure 22.58(b), using the rules of vector addition to sum the contributions from each wire.

To determine

The magnitude and the direction of the magnetic field at the point equidistant from the wires.

Explanation

Given info:

Current through wire A is,IA=10.0â€‰A.

Current through wire B is,IB=10.0â€‰A.

Current through wire C is,IC=5.0â€‰A.

Current through wire C is,ID=5.0â€‰A.

Formula used:

General formula to calculate the magnetic field produced by a long straight wire is given as,

B=Î¼0I2Ï€râ€ƒâ€ƒâ€ƒâ€ƒâ€ƒâ€ƒâ€ƒ...... (1)

Calculation:

Consider the given figure.

The equidistant point will be the center of the square formed by four wires that is pointO.

The distance of pointOfrom each wire is,

r=120.202+0.202r=14.14Ã—10âˆ’2â€‰m

Substituting the values from the figure for wire A in equation (1), we get

BA=(4Ï€Ã—10âˆ’7â€‰Tâ‹…m/A)(10.0â€‰A)2Ï€(14.14Ã—10âˆ’2â€‰m)BA=1.414Ã—10âˆ’5â€‰T

Substituting the values from the figure for wire B in equation (1), we get

BB=(4Ï€Ã—10âˆ’7â€‰Tâ‹…m/A)(10.0â€‰A)2Ï€(14.14Ã—10âˆ’2â€‰m)BB=1.414Ã—10âˆ’5â€‰T

Substituting the values from the figure for wire C in equation (1), we get

BC=(4Ï€Ã—10âˆ’7â€‰Tâ‹…m/A)(5

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