I want to use Faradays law to convert the kinetic energy from actions such as a jog/shake to produce electricity to ultimately charge a battery.


  • Neodynium Magnet 3/4 diameter (2)
  • Solenoid casing (x2)
  • Enameled Copper wire (x2)
  • 1800mAh battery
  • Spring (x2)


2 solenoids with the magnet inside connected to a spring that will move up and down or use urethane stoppers. Then use a rectifier to convert AC to DC and charge the battery.

So I'm essentially adopting the faraday flashlight concept using two solenoids. How do I calculate current output and how do I to go about increasing efficiency?

  • 2
    \$\begingroup\$ What is your question? Also, adding some paragraph breaks to your post would improve it immensely. \$\endgroup\$
    – uint128_t
    Feb 9 '16 at 1:14
  • \$\begingroup\$ I tried adding the breaks the mobile version crunches everything together, my question is would that be a practical voice of design If I were to use a 1800Mah battery 3.6V, with two solenoids and two neodynium magnets cylindrical in nature. With the magnets vibrating back and forth \$\endgroup\$
    – Shane
    Feb 9 '16 at 1:44

Sure, a magnet bouncing through a coil will generate power. You can see plenty of LED flashlights that work like this.

An 1800mAh - I assume you didn't mean mega-amp-hour ;) - battery is pretty big though, and it would take many hours of shaking to charge one. Super-rough calculation: a 20g magnet moving at 2m/s has 40mJ of kinetic energy, and lets say you managed to recover 50% of that, for 20mJ of electricity per stroke. 1800mAh at 3.6V for a Li-Ion battery is 23.3kJ so you need 1.2 MILLION strokes to charge the battery. Say it is bi-directional and you can shake it at 5Hz, that's producing 20mJ pulses at 10Hz... so 32.4 hours of constant vigorous shaking to charge the battery.

The voltage you get will be a function of the magnetic field strength (how good is your magnet), the speed of the magnet through the coil and the number of windings on the coil. Refer to Maxwell's equations...

The current will be approximately the voltage produced by the coil, less the voltage of the battery you are trying to charge, divided by the total impedance of everything (coil, battery, etc) in that loop. It might also be limited by the inductance of your coil.

The current so produced will cause a dragging force on the magnet. If the current is too small, the magnet will whizz straight through the coil and produce no power, but if the current is too large, the magnet will stall in the coil. You will need to design the coil and charging circuit in relationship with the mass of the moving magnet so that it extracts a reasonable quantity of kinetic energy per motion.

  • \$\begingroup\$ I appreciate the detail that you put in the answer, so one of many things is that if I increased the amount of Turns (N) and have two solenoids on each side with the battery in middle and if I decrease the length of the solenoids wouldn't that in practice help with the shaking or use springs with a more suitable spring constant. I'm using my college level physics education so sorry for being a newbie. \$\endgroup\$
    – Shane
    Feb 9 '16 at 20:45

For comparison with William Brodie-Tyrrell's excellent answer:

1800 mAh at 3.6 V = 6.48 Wh (watt-hours).

A reasonably fit person can keep up > 100 W on a bike for a few hours so I imagine one could hand crank a 5 W generator without too much trouble. Time taken to hand-charge the battery = 6.5 / 5 = 1.3 hours.


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.