General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 20, Problem 44E

(a)

To determine

The magnitude of flux due to the current in the strip.

(a)

Expert Solution
Check Mark

Answer to Problem 44E

The elemental flux through the strip is 2ckirdr.

Explanation of Solution

Write the expression for flux in the through strip.

    dϕ=BdA        (1)

Here, dϕ is the small amount of flux, B is the magnetic field and dA is the elemental area.

Write the expression for the magnetic field in the strip.

    B=2kir

Here, k is the magnetic constant, i is the current in the wire and r is the distance of the strip from the wire.

Write the expression for the area of the strip.

    dA=cdr

Here, c is the length of the loop and dr is small elemental width.

Conclusion:

Substitute 2kir for B and cdr for dA in equation (1).

    dϕ=(2kir)(cdr)=2ckirdr

Thus, the elemental flux through the strip is 2ckirdr.

(b)

To determine

The magnitude of the flux through the loop.

(b)

Expert Solution
Check Mark

Answer to Problem 44E

The flux through the loop is 2cki[ln(baa)].

Explanation of Solution

Write the expression for the elemental flux through the strip in the loop.

    dϕ=(2kir)(cdr)=2ckirdr        (2)

Conclusion:

Integrate equation (2).

    dϕ=2ckiabadrr

Here, a is the distance of the first side of the loop from the wire and b is the distance of the last side of the loop from the wire.

Solve the above integration.

    ϕ=2cki[ln(ba)ln(a)]=2cki[ln(baa)]

Thus, the flux through the loop is 2cki[ln(baa)].

(c)

To determine

The magnitude of the induced EMF through the loop.

(c)

Expert Solution
Check Mark

Answer to Problem 44E

The induced EMF through the loop is (2ck[ln(baa)]didt).

Explanation of Solution

Write the expression for induced EMF.

    ε=dϕdt        (3)

Here, ε is the induced EMF.

Write the expression for the flux through the loop.

    ϕ=2cki[ln(baa)]        (4)

Conclusion:

Differentiate equation (4) with respect to t.

    dϕdt=2ck[ln(baa)]didt

Here, didt is the change in current and dϕdt is the change in flux.

Substitute (2ck[ln(baa)]didt) for dϕdt in equation (3).

    ε=2ck[ln(baa)]didt

Thus, the induced EMF through the loop is (2ck[ln(baa)]didt).

(d)

To determine

The induced current in the loop.

(d)

Expert Solution
Check Mark

Answer to Problem 44E

The induced current in the loop is (2cRk[ln(baa)]didt).

Explanation of Solution

Write the expression for induced EMF in the loop.

    ε=2ck[ln(baa)]didt        (5)

Write the expression for the induced current in the loop.

    i1=εR        (6)

Here, i1 is the induced current and R is the resistance of the loop.

Conclusion:

Substitute (2ck[ln(baa)]didt) for ε in equation (6).

    i1=2cRk[ln(baa)]didt

Thus, the induced current in the loop is (2cRk[ln(baa)]didt).

(e)

To determine

The direction of i1 for increasing in i..

(e)

Expert Solution
Check Mark

Answer to Problem 44E

The direction of the induced current is clockwise.

Explanation of Solution

Write the expression for i1.

    i1=2cRk[ln(baa)]didt

Conclusion:

The induced current is proportional to the change in i.

Thus, the direction of the induced current is clockwise.

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Chapter 20 Solutions

General Physics, 2nd Edition

Ch. 20 - Prob. 11RQCh. 20 - Prob. 12RQCh. 20 - Prob. 13RQCh. 20 - Prob. 1ECh. 20 - Prob. 2ECh. 20 - Prob. 3ECh. 20 - Prob. 4ECh. 20 - Prob. 5ECh. 20 - Prob. 6ECh. 20 - Prob. 7ECh. 20 - Prob. 8ECh. 20 - Prob. 9ECh. 20 - Prob. 10ECh. 20 - Prob. 11ECh. 20 - Prob. 12ECh. 20 - Prob. 13ECh. 20 - Prob. 14ECh. 20 - Prob. 15ECh. 20 - Prob. 16ECh. 20 - Prob. 17ECh. 20 - Prob. 18ECh. 20 - Prob. 19ECh. 20 - Prob. 20ECh. 20 - Prob. 21ECh. 20 - Prob. 22ECh. 20 - Prob. 23ECh. 20 - Prob. 24ECh. 20 - Prob. 25ECh. 20 - Prob. 26ECh. 20 - Prob. 27ECh. 20 - Prob. 28ECh. 20 - Prob. 29ECh. 20 - Prob. 30ECh. 20 - Prob. 31ECh. 20 - Prob. 32ECh. 20 - Prob. 33ECh. 20 - Prob. 34ECh. 20 - Prob. 35ECh. 20 - Prob. 36ECh. 20 - Prob. 37ECh. 20 - Prob. 38ECh. 20 - Prob. 39ECh. 20 - Prob. 40ECh. 20 - Prob. 41ECh. 20 - Prob. 42ECh. 20 - Prob. 43ECh. 20 - Prob. 44ECh. 20 - Prob. 45ECh. 20 - Prob. 46ECh. 20 - Prob. 47ECh. 20 - Prob. 48ECh. 20 - Prob. 49ECh. 20 - Prob. 50ECh. 20 - Prob. 51ECh. 20 - Prob. 52ECh. 20 - Prob. 53ECh. 20 - Prob. 54ECh. 20 - Prob. 55ECh. 20 - Prob. 56ECh. 20 - Prob. 57ECh. 20 - Prob. 58ECh. 20 - Prob. 59ECh. 20 - Prob. 60ECh. 20 - Prob. 61ECh. 20 - Prob. 62ECh. 20 - Prob. 63ECh. 20 - Prob. 64ECh. 20 - Prob. 65ECh. 20 - Prob. 66ECh. 20 - Prob. 67ECh. 20 - Prob. 68ECh. 20 - Prob. 69ECh. 20 - Prob. 70ECh. 20 - Prob. 71ECh. 20 - Prob. 72ECh. 20 - Prob. 73ECh. 20 - Prob. 74ECh. 20 - Prob. 75ECh. 20 - Prob. 76ECh. 20 - Prob. 77ECh. 20 - Prob. 78ECh. 20 - Prob. 79ECh. 20 - Prob. 80ECh. 20 - Prob. 81ECh. 20 - Prob. 82ECh. 20 - Prob. 83ECh. 20 - Prob. 84ECh. 20 - Prob. 85ECh. 20 - Prob. 86ECh. 20 - Prob. 87ECh. 20 - Prob. 88ECh. 20 - Prob. 89ECh. 20 - Prob. 90ECh. 20 - Prob. 91ECh. 20 - Prob. 92ECh. 20 - Prob. 93ECh. 20 - Prob. 94ECh. 20 - Prob. 95E
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