Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 30, Problem 20P

(a)

To determine

The proof that B=μ0Ia2πR1R2+a2 .

(a)

Expert Solution
Check Mark

Answer to Problem 20P

It is proved that B=μ0Ia2πR1R2+a2 .

Explanation of Solution

Formula used:

The expression for magnetic field is given by,

  B=μ0Ia4πR(sinθ1+sinθ2)

The expression for angle is given by,

  sinθ=aR2+a2

Calculation:

The magnetic field is calculated as,

  B=μ0Ia4πR(sinθ1+sinθ2)=μ0Ia4πR(sinθ+sinθ)=μ0Ia4πR(a R 2 + a 2 +a R 2 + a 2 )=μ0Ia2πR1 R 2 + a 2

Conclusion:

Therefore, it is proved that B=μ0Ia2πR1R2+a2 .

(b)

To determine

The proof that ExdA=Q0( r 2 + a 2 )32πdr .

(b)

Expert Solution
Check Mark

Answer to Problem 20P

It is proved that ExdA=Q0( r 2 + a 2 )32πdr .

Explanation of Solution

Formula used:

The expression for electric field is given by,

  Ex=14π02Qa( r 2 + a 2 )32

The expression for flux is given by,

  ExdA=Ex(2πrdr)

Calculation:

The expression for flux is calculated as,

  ExdA=Ex(2πrdr)=14π02Qa ( r 2 + a 2 ) 3 2 (2πrdr)=Qa0 ( r 2 + a 2 ) 3 2 rdr=Q0 ( r 2 + a 2 ) 3 2 πdr

Conclusion:

Therefore, it is proved that ExdA=Q0( r 2 + a 2 )32πdr .

(c)

To determine

The proof that ϕe=Q0(1a a 2 + R 2 ) .

(c)

Expert Solution
Check Mark

Answer to Problem 20P

It is proved that ϕe=Q0(1a a 2 + R 2 ) .

Explanation of Solution

Formula used:

The expression for flux from (b) is given by,

  0dϕe=Qa0( r 2 + a 2 )32(rdr) …… (1)

Calculation:

Integrate equation (1) from 0 to R .

  dϕe=100R Qa 0 ( r 2 + a 2 ) 3 2 ( rdr)=Qa0( 1 R 2 + a 2 +1a)=Q0(1a R 2 + a 2 )

Conclusion:

Therefore, it is proved that ϕe=Q0(1a a 2 + R 2 ) .

(d)

To determine

The proof that I+Id=Iaa2+R2 .

(d)

Expert Solution
Check Mark

Answer to Problem 20P

It is proved that I+Id=Iaa2+R2 .

Explanation of Solution

Formula used:

The expression for displacement current is given by,

  Id=I(1a R 2 + a 2 )

Calculation:

The summation of current and displacement current is calculated as,

  I+Id=I+(I( 1 a R 2 + a 2 ))=II(1a R 2 + a 2 )=II+Ia R 2 + a 2 =Ia R 2 + a 2

Conclusion:

Therefore, it is proved that I+Id=Iaa2+R2 .

(e)

To determine

The proof that result from generalized ampere’s law is similar to part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 20P

It is proved that result from generalized ampere’s law is similar to part (a).

Explanation of Solution

Formula used:

The expression for generalized form of ampere’s law is given by,

  Bdl=μ0(I+Id)

Calculation:

The magnetic field is calculated as,

  Bdl=μ0(I+Id)B(2πR)=μ0Ia R 2 + a 2 B=μ0I2πRa R 2 + a 2 =μ0Ia2πR1 R 2 + a 2

Conclusion:

Therefore, it is proved that result from generalized ampere’s law is similar to part (a).

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1.62 On a training flight, a Figure P1.62 student pilot flies from Lincoln, Nebraska, to Clarinda, Iowa, next to St. Joseph, Missouri, and then to Manhattan, Kansas (Fig. P1.62). The directions are shown relative to north: 0° is north, 90° is east, 180° is south, and 270° is west. Use the method of components to find (a) the distance she has to fly from Manhattan to get back to Lincoln, and (b) the direction (relative to north) she must fly to get there. Illustrate your solutions with a vector diagram. IOWA 147 km Lincoln 85° Clarinda 106 km 167° St. Joseph NEBRASKA Manhattan 166 km 235° S KANSAS MISSOURI
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