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## Main Question or Discussion Point

If a permanent magnet rod enters a coil of wires at a certain speed will the permanent magnet slow down? I'm trying to build a generator of some sort. So i need to know is my logic sound. I'll give you scenario.

Scenario:

The permanent magnet slides into the coil at a speed of 3meters/second. What will the speed of the magnet be after exiting the coil?

Permanent magnet parameters

Mass : 2kg

Strength : 2000Gauss or 0.2Tesla

Area : 0.008m2

Wire Coil Parameters

Turns : 20000

Resistance : 50 Ohm

Length of Coil : 3meters

Thus the energy contain in the magnet moving at 3ms-1 is = 9joules

E=1/2 mv^2

E=(1/2)(2)(3^2)

E=9Joules

So using faradays law, the power generated by the magnet passing thru the coil of wire :

V= (-N)(BA/t)

V = voltage

N = Turns

B = Flux (in tesla)

t = Round per second

A = Surface Area of permanent magnet

Note : I will reduce the flux strength to 0.1tesla because the coil will be approximately 2mm from the magnet

V = (-20000)(0.1*0.008*/1)

V = 16Volts

Thus the current generated is

I=V/R

I = 16V / 50ohm

I = 0.32amps

Thus power generated is

Power = VI

Power = 16V * 0.32amps

Power = 5.12Watts or joule per second

Thus the speed of the permanent magnet exiting the coil is

9joules – 5.12 joules = Kinetic Energy Remaining

3.88 = (1/2)(2)(v ^2)

v = √(3.88*2/2)

v = 1.96ms-1

Is this logic and calculation correct??

Scenario:

The permanent magnet slides into the coil at a speed of 3meters/second. What will the speed of the magnet be after exiting the coil?

Permanent magnet parameters

Mass : 2kg

Strength : 2000Gauss or 0.2Tesla

Area : 0.008m2

Wire Coil Parameters

Turns : 20000

Resistance : 50 Ohm

Length of Coil : 3meters

Thus the energy contain in the magnet moving at 3ms-1 is = 9joules

E=1/2 mv^2

E=(1/2)(2)(3^2)

E=9Joules

So using faradays law, the power generated by the magnet passing thru the coil of wire :

V= (-N)(BA/t)

V = voltage

N = Turns

B = Flux (in tesla)

t = Round per second

A = Surface Area of permanent magnet

Note : I will reduce the flux strength to 0.1tesla because the coil will be approximately 2mm from the magnet

V = (-20000)(0.1*0.008*/1)

V = 16Volts

Thus the current generated is

I=V/R

I = 16V / 50ohm

I = 0.32amps

Thus power generated is

Power = VI

Power = 16V * 0.32amps

Power = 5.12Watts or joule per second

Thus the speed of the permanent magnet exiting the coil is

9joules – 5.12 joules = Kinetic Energy Remaining

3.88 = (1/2)(2)(v ^2)

v = √(3.88*2/2)

v = 1.96ms-1

Is this logic and calculation correct??