Difference between revisions of "Chapter 29 Problem 35"

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__NOTOC__
 
==Problem==
 
==Problem==
 
[[File:Chapter29Problem35q.png|300px|center|Wires and currents]]
 
[[File:Chapter29Problem35q.png|300px|center|Wires and currents]]
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(b) in the opposite direction to <math>I</math>.
 
(b) in the opposite direction to <math>I</math>.
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==Solution==
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[[File:Chapter29Problem35s.png|300px|center|Wire and current induced magnetic field]]
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The magnetic field is perpendicular to the rod, with the magnetic field decreasing with distance from the rod.
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===(a)===
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<math>d\mathcal{E}=Bvdr</math>
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<math>B=\frac{\mu_0}{2\pi}\frac{I}{r}</math> (near a straight long wire)
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<math>\mathcal{E}=\int d\mathcal{E}=\int_b^{b+a}\frac{\mu_0 I}{2\pi r}v dr=\frac{\mu_0 Iv}{2\pi }\ln\left(\frac{b+a}{b}\right)</math>
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The force is towards the long wire
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===(b)===
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same magnitude, away from the long wire

Latest revision as of 22:41, 10 May 2020

Problem

Wires and currents

A short section of wire, of length , is moving with velocity , parallel to a very long wire carrying a current . The near end of the wire section is a distance from the long wire. Assuming the vertical wire is very long compared to , determine the emf between the ends of the short section. Assume is

(a) in the same direction as ,

(b) in the opposite direction to .

Solution

Wire and current induced magnetic field

The magnetic field is perpendicular to the rod, with the magnetic field decreasing with distance from the rod.

(a)

(near a straight long wire)

The force is towards the long wire

(b)

same magnitude, away from the long wire