Electronic – Electromagnet for string vibration

You are assuming that the sensor in the road detects a magnet (it doesn't), or that it detects ferrous metals like iron (also, negative).

A 20 kg electromagnet might make it easier for your bike to trigger the sensor, but not because it is an electromagnet-- because you have added 20 kg of metal to your bike.

Here is a nice web page that talks about road sensor designs and how you might get your bike to be more easily sensed. Essentially, you place your bike in strategic locations on the sensor or buy a car. One note about that page: it talks about mounting permanent or electro-magnets to your bike, but that won't work well if at all. Positioning your bike is much more effective.

Here is a formula that is useful

Force = \$(N\cdot I)^2\cdot 4\pi 10^{-7}\cdot \dfrac{A}{2g^2}\$

• F = Force
• I = Current
• N = Number of turns
• g = Length of the gap between the solenoid and the magnetizable metal
• A = Area

With 100 amps through a 100 turn coil of area 1 sq m, the force on a magnetizable metal at 100mm is 6283 newtons (about 1400 pounds).

But what about the permeability of the electromagnet's core and the permeability of the "iron" it's attracting to - air forms the gap and it dictates the flux density.

The input power needed to do this is zero because no work is being done in creating this force. However, if you actually want to know the losses you need to think about how thick the copper wire is to pass this current and maybe trade off number of turns with amps.

Here's an online calculator and here is a page that explains the theory.