Physics Homework Problem
A child (approximately 4 years old) takes
her metal "Slinky Toy"
and does various tests to determine
that the Slinky has an inductance 125 H
when it has been stretched to a length of 3 m.
The permeability of free space is 4pi
10x e-7, N=A2.
If a slinky has a radius of 4 cm, what is the
total number of turns in the Slinky?
Now that's a mest up kid.
posted by Katherine @ 6:58 p.m.
2 Comments:
Lols.
I heart Kat.
Let's see.
Inductance can be written as
L=[(permeability of freespace)*(relative permeability of core)*(Cross sectional area of coil)*(# of turns [N])^2]/ (length of coil in meters)
Permeability of Freespace= 4*3.14*10e-7
And since slinkys were made of either steel of aluminum, for this we'll assume that it is made of steel which has a permeability of of 875 microNetwons per Amp squarred.
Realtive permeability of core= Permeability of substance/Perm of Free space.
In this case terms cancel and we are left with
L=[(Permeability of Steel)*(# of turns)^2 * (cross sectional area)]/length
S=cross sectional area
125H=[(875e-6)*(N^2)*pi*.04^2]/3
solve for N=# of turns of wire.
btw, N=A2 didn't really make sense to me so i disregarded that. and i assumed it was a steel slinky, for all i know it's aluminum.
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