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onoi
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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??