magnetorque

can any one help me ....i have cal ulate the parameter for the magnetorque.....but it result mpre more than i expect...


suggest me please
here is calculation
Length of iron ferrite core=80 mm
Radius of core=6 mm
Power=1 watt
Voltage=5 volt


Firstly,expression for the minimum resistance
R= v^2/P
= 25/1
= 25 ohm

Then,
Length of wire (L)= Resistance/resistivity of wire (copper)
=25/1.55×××××*10^-8
=1612903226

Then,

No.of turns. (N)=Resistance/2Πr* resistivity of wire
=25/2*3.1428*1.55*10^-8
=132578168

THEN,

Demagnetizing factor= 4[Ln(l/r)-1]/(l/r)^2-4 ln(l/r)
As putting the value of length of core and radius of core,the result comes as 0.01306274



Then finally for magnetic dipole moment


M= r v/2*resivity of wire[1+relative permebility of core -1/1+ (MIU-1) Nd]


It results as 36.15*10^9 am^2



Resistivity of copper wire =1.55*10^-8 ohm metre.

What is The Magnetorque?

Your calculations for wire length and number of turns don't take into account the cross-sectional area of the wire.
You should specify your units of measurement.

can you calculate please

We don't do your homework for you, but can give guidance.
First of all you need to give us the equation for wire resistance (in Ohms) as a function of resistivity, wire radius and wire length, specifying the units used.
We still don't know what you mean by 'magnetorque'. Google gives various definitions, none of which seem relevant to your problem.

Magnetorque refers to a magnetic clutch with coupling of a rotating motor.
So please call this .... a magnetic force like a linear solenoid or linear voice coil.

Resistivity of copper wire =1.55*10^-8 ohm metre
This is not R but refers to conductivity, ρ where resistance factors length ,L over cross-sectional area , A
R= ρ L/A

silver is better 1.59 and copper is 1.68 *10^-8 ohm metre not 1.55

if core radius , r=6 mm or = 2π*r = 37.7 mm per turn

One way is to use look up table for R instead of ρ.

A change in Ten (10) gauge numbers, for example from No. 30 to 40, reduces the area, A and weight by ~ /10 and increases R~ x10
Resistance approximately doubles every AWG guage size. (ref WIKI)

AWG30 will fuse open with 10A in 10 seconds so 5% of that is acceptable or 0.5A and you require 0.2A
AWG30 has an area of 0.0509 mm^2 and R= 338.6 Ω/m thus 25Ω length is 74 mm or N=2 turns

AWG36 is probably too small wire size and may be too warm with 1 Watt dissipate
but A= 0.0127 mm^2 and R= 1361 mΩ/m but 25Ω is 18+ turns

You can do the rest.