I'am given this picture with the following information enter image description here

Core's length L=1.8m cross-section area A=(15x17)cm^2 gap δ=5mm winding turns w=72 winding resistance R=150mΩ, μr=1800, i(t)=25sin100π I can answer questions B and C but they require finding question A which asks what is the magnetic field B. Faraday's law comes to mind but we don't have u(t) and also Magnetic potential $$V=LH_{core}+δH_0$$ But again I don't have V.


Current is a peak of 25 amps and, you have the turns (72) so, you have the MMF (magneto motive foce) because that equals amps x turns. So MMF = 1800 (peak).

You also have the length of the core and you can reasonably assume that 95%+ of the magnetic field flows through the core and gap.

Magnetic field strength (H) = MMF/length of core. Hence, H = 1000 ampere turns per metre. Looks like the numbers are working out nicely.

Next you have to work out effective permeability i.e. the permeability with the gap and this formula is as shown on this calculator: -

enter image description here

You need to multiply this by the permeability of a vacuum (air) to get a real number so, a relative permeability of 300 becomes an absolute permeability of 0.000377.

B = H x permeability hence B (peak) = 0.377 teslas.

Area and resistance (although quoted in the question), are not used in this calculation.

|improve this answer|||||
  • \$\begingroup\$ I prefer to think of it this way: energy cannot be stored in solid state matter, only vacuum. (Quantum spin of matter can get in the way, so to speak, and must be discounted.) \$\mu_r\$ represents the relative amount of vacuum found in the physical solid state matter of the core material. So this makes your equation very easy to remember: \$L_{tot}=L_{gap}+\frac{L_{core}}{\mu_r}\$, which says the total vacuum path length is the sum of the gap you have plus the gap that exists inside the solid state matter. As always then, \$\mu_{eff}=\frac{L_{core}}{L_{tot}}\$. Some algebra gets your equation. \$\endgroup\$ – jonk Sep 5 '17 at 20:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.