Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305714892
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 30, Problem 30.19P

Determine the magnetic field (in terms of I, a, and d) at the origin due to the current loop in Figure P29.9. The loop extends to infinity above the figure.

Figure P29.9

Chapter 30, Problem 30.19P, Determine the magnetic field (in terms of I, a, and d) at the origin due to the current loop in

Expert Solution & Answer
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To determine
The magnetic field at the origin due to current loop.

Answer to Problem 30.19P

The magnetic field at the origin due to current loop is μoI2πad[d2+a2d] .

Explanation of Solution

Given Info: current flowing through the loop is I , distance of point O from loop is d and length of horizontal part of current loop from origin is a .

Explanation:

Formula to calculate magnetic field given by Biot savarts law,

dB=μoI4πdlsinθr2r (1)

Formula to calculate r from Pythagorean theorem,

r=a2+d2

Write the expression to calculate r from the figure (1),

cosecθ=rar=acosecθ

Similarly,

tanθ=all=acotθ

Differentiate above equation,

dl=d(acotθ)=a[d(cotθ)]=a(cosec2θ)dθ=acosec2θdθ

Substitute acosec2θdθ for dl and acosecθ for r in equation (1)

dB=μoI4π(acosec2θdθ)sinθ(acosecθ)2r^dB=μoI4πasinθdθr^ (2)

On integrate both sides from limit θ1toθ2 .

dB=θ1θ2μoI4πasinθdθrB=μoI4πaθ1θ2sinθdθr=μoI4πa[cos]θ1θ2=μoI4πa[cosθ1cosθ2] (3)

For length left to point O , θ2=0 and cosθ1=dd2+a2 .

BOA=μoI4πa[dd2+a2cos0]=μoI4πa[dd2+a21]

For length between O and loop, cosθ1=ad2+a2 and cosθ2=ad2+a2 .

BAB=μoI4πd[ad2+a2+ad2+a2]

For length right to point O , cosθ1=dd2+a2 and cosθ2=180 .

BBC=μoI4πa[cos180(dd2+a2)]=μoI4πa[1+dd2+a2]

Net magnetic field at point O is,

BO=BOA+BAB+BBC

Substitute μoI4πa[1+dd2+a2] for BOA , μoI4πd[ad2+a2+ad2+a2] for BAB and μoI4πa[dd2+a21] for BBC in above equation.

BO=μoI4πa[1+dd2+a2]+μoI4πd[ad2+a2+ad2+a2]+μoI4πa[+dd2+a21]=μoI4π[1adad2+a2]+[add2+a2+add2+a2]+[dad2+a21a]=μoI4π[dd2+a2+d2+2a2+d2dd2+a2add2+a2]=μoI2π[d2+a2dd2+a2add2+a2]=μoI2πad[d2+a2d]

Conclusion:

Therefore, magnetic field at point O is μoI2πad[d2+a2d] .

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Chapter 30 Solutions

Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term

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