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Introductory Physics Volume Two

Introductory Physics Volume Two

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110 Sources of Magnetic Field 5.4<br />

since d ⃗ l and the radius vector to the point are perpendicular. The<br />

length of dl can be approximated by the length of the arc subtended<br />

by the angle ∆θ, so<br />

|dB| ⃗ = Iµ 0 (adθ)(a)<br />

4π a 3 ,<br />

= Iµ 0 dθ<br />

4π a<br />

To find the magnetic field due to the entire wire, integrate:<br />

B = Iµ 0<br />

4π<br />

∫ π<br />

0<br />

dθ<br />

a = Iµ 0<br />

4a<br />

Example<br />

<strong>Two</strong> sections of straight wire carrying electric current lie parallel to<br />

each other, a distance b apart:<br />

I 2<br />

b<br />

a<br />

I 1<br />

The leftmost ends of the wires are aligned. What is the force that<br />

current I 1 exerts on current I 2 ? Draw a coordinate system:<br />

y<br />

dx'<br />

x'<br />

I 2<br />

b<br />

I 1<br />

x<br />

x=-<br />

a<br />

2<br />

The z-axis points out of the page. From a previous example, we know<br />

that the magnetic field due to I 1 in the xy-plane is<br />

⎛<br />

⎞<br />

⃗B 2 (x, y) = µ 0I 1 x +<br />

⎝<br />

a 2<br />

x − a 2<br />

√<br />

4πy (x )<br />

− √ ⎠ ˆk<br />

+<br />

a 2<br />

2 + y<br />

2 (x −<br />

a<br />

2 )2 + y 2

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