If you know the current? Magnetic fields confuse me a bit in general and I‘ve googled this numerous times and haven‘t been able to find a simple answer. Please help!
Solenoid Magnetic Field Calculator
H nI. It's as simple as that! E.g. a 20cm long solenoid has 500 windings along its length and carries a current of 2A n 500/0.2 2500 turns/metre H 2500 x 2 5000 A/m (inside the solenoid, not near the ends) ___________________ However, we often need the flux density, B (units tesla). If there is a vacuum or air in the coil, then B μ0H where μ0 is the permeability of free space (4π x10^-7 H/m). But if there is a magnetic material in the coil (e.e.g iron) then this changes B (but H is not effected - H is the magnetic field in the absence of the material). To find B, we use B μr.μ0H where μr is the relative permeability of the material. Sometimes μr and.μ0 are combined into a single constant, μ (μr.μ0), which is just called the permeability of the material. (The factor 4π x10^-7 sometimes causes confusion. it arises because of the way various quantities are defined in the SI system, If you were working in the old-fashioned cgs system of units, the unit for H is the 'oerstead' and the unit for B is the gauss. And 1 oerstead is equivalent 1 gauss, so it is easier to follow the B-H calculations.) Not sure if that helps though!
by increasing the current or increasing the turn no. of the coil u can increase the field strength produced by a coil
