Physics help - magnetism
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Physics help - magnetism

[From: ] [author: ] [Date: 11-05-06] [Hit: ]
Assume the horizontalcomponent of the Earths field at this point is 0.45x10-4T and the magnetic declination is 0 degrees.of the earth (the true north) …… http://en.wikipedia.The needle of a compass always lies along the magnetic field lines of the earth.If the magnetic declination at a point on the earth’s surface is zero,......
Even with the examples from the book, I'm confused by how to set this up and exactly what I'm solving for. Can anyone explain at all?

A horizontal compass is placed 18 cm due south from a straight vertical wire carrying 35 A current downward. In what direction does the compass needle point at this location?Assume the horizontal component of the Earth's field at this point is 0.45x10-4T and the magnetic declination is 0 degrees.

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Magnetic declination is the deviation of the magnetic north (the direction
the north end of a compass needle points) from the geographic north pole
of the earth (the true north) …… http://en.wikipedia.org/wiki/Magnetic_de…
The needle of a compass always lies along the magnetic field lines of the earth.
If the magnetic declination at a point on the earth’s surface is zero, it means that
the horizontal component of the earth’s magnetic field line at that point lies along
the line of the north-south magnetic poles.
Now, the presence of a current carrying straight vertical wire creates an additional
magnetic field that must be combined with the earth’s magnetic field. Since magnetic
fields are vectors, the magnetic field of the earth and the magnetic field of the vertical
wire must be combined vectorially.
With the magnetic field of the earth along the +x-axis (B₁ = 0.45 × 10 ⁻ ⁴ T along the
south-north magnetic pole line) and the magnetic field due to the straight vertical
wire along the +y-axis (B₂ along the east-west direction obtained by using the right
hand rule of Ampere’s law), where ……
…… B₂ = μ₀ i / [ 2 π R ] = [ 4π × 10 ⁻ ⁷ T • m / A ] ( 35 A ) / [ 2 π (18 × 10 ⁻ ² m ) ]
………. = 3.9 × 10 ⁻ ⁵ T = 0.39 × 10 ⁻ ⁴ T
we find that …… tan θ = B₂ / B₁ = 0.39 × 10 ⁻ ⁴ T / 0.45 × 10 ⁻ ⁴ T ……
which gives θ = 41°.
The compass needle points along the direction of 41° west of north.
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