Question

If you pass a current through a wire, you can create a magnetic
field. In this class we examine the field from three specific
configurations: an infinitely-long straight wire, a compact loop of
wire (possibly with multiple turns, a coil), or an infinitely long
solenoid (think of an infinitely-long cylinder of loops made by
wrapping a single wire). See the text for an extended description
of all three geometries.

Here’s the challenge: there are three slightly different formulas
used to calculate the magnetic field. Which you use depends on the
configuration of the wire. Look in the text, check out the figures,
and write down the formula for each of these three situations in
your notes. Label the formulas and read carefully to make sure you
understand what each variable represents.

Suppose you have an infinitely long wire carrying a current of 19.0
A. What is the strength of the magnetic field a distance 0.0360 m.
away from the wire?

You are now studying a length of wire that has been coiled up to create a circular coil of 8 turns (individual loops) all arranged compactly. The loop has radius of 0.0270 m and a current of 29.4 A. is passing through the wire. What is the strength of the magnetic field at the center of the loop?

Lastly, you are handed a long solenoid. The special
characteristic of an infinitely-long solenoid is that the magnetic
field inside the solenoid is uniform. There are 4.20×10^{3}
turns over a length of 0.110 m and a current of 2.75 A flows
through the wire. What is the strength of the magnetic field inside
the solenoid?

Answer #1

Magnetic field due to straight wire, B_{1} = _{o}
i/2d

= 4*10^{-7}*19 / 2*0.036

=1.055*10^{-4} T

Magnetic field due to a coil, B_{2} = _{o}
Ni/2R

= 4**10^{-7}*29.4*8
/ 2*0.027

=5.47*10^{-3} T

Magnetic field due to infinite solenoid, B_{3} =
_{o}
Ni/L

= 4**10^{-7}*2.75*4.2*10^{3}/
0.11

=0.13188 T

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