An inductor is to be made by wrapping wires around a paper tube of radius 0.702 cm and length 8.44 cm. How many turns will give the inductor an inductance of 8.000e-3 H?I thought this would be fairly easy but I think i am either using the wrong Area formula or confusing the number of turns with the number of loops calculation.I used the formula: L n^2 * s * mu * A where n is the number density of loops where n N/sN number of loops, s Length. mu is a constant (4*pi) * 10^-7 or 1.26E-6. The area i used was the Area of a Cylinder since the question asked for a paper tube so the area of an open ended cylinder is A 2*pi*r*L, where r radius and L LengthI solved for N where N SQRT[ (L*s) / (mu*A) ].So i got N 379 loops.and i dont know if this is the same thing as turns but if its not, please explain what it is asking for and show a derivation to the final answer as well as the final answer.thank you!PLEASE HELP!!!
Set the length of the lighter material to x and the other to 16-x length*area*density+length*area*densit. x*pi*7^2*4.3+(16-x)*pi*7^2*6.313103 x7.8cm
You have it right. Turns and loops are the same thing. 8mH is quite a large inductor. I haven't checked your calculations but by experience I'll tell you that if you ever wind your own coils it takes forever. So nearly 400 loops or turns - yes that doesn't sound wrong.
An inductor is to be made by wrapping wires around a paper tube of radius 0.702 cm and length 8.44 cm. How many turns will give the inductor an inductance of 8.000e-3 H? L μ 0 * N^2 * A ÷ length μ 0 4π * 10^-7 T * m /A L 8.0* 10^-3 Length 8.44 cm 8.44 * 10^-2 m Radius 0.702 cm 7.02 * 10^-3 m Area of circle π * r^2 π * (7.02 * 10^-3)^2 1.548 * 10^-4 m^2 8.000 * 10^-3 (4π * 10^-7) * N^2 * (1.548 * 10^-4) ÷ (8.44 * 10^-2) Multiply both sides by (8.44 * 10^-2) (8.44 * 10^-2) * (8.000 * 10^-3) (4π * 10^-7) * N^2 * (1.548 * 10^-4) N^2 [(8.44 * 10^-2) * (8.000 * 10^-3)] ÷ [(4π * 10^-7) * (1.548 * 10^-4)] N^2 3.47 * 10^6 N 1862 turns L μ 0 * N^2 * A ÷ length
