current & force fomulas

searley · May 21, 2009

got these questions of a pal and can`t work out the formulas but got the answers to the questions.

1,a conductor 500mm long is placed in at a right angle to a meagnetic flux density of 0.25 tesla,what is the current required in the conductor to create a force of 15N on the concuctor = 120A but what is the formula ?

2, a conductor 300mm long is placed in at a right angle to a magnetic flux density of 0.3 tesla,what is the fotce on the conductor if a current of 36A is passed through it =3.24N but what is the formula ?

3,the air gap in a conductor is 25mm2,the flux density is 1.5 TESLA.calc the total flux = 0.0375mwb but what is the formula ?

thanks for the help.

msb2003 · May 21, 2009

Questions 1 & 2 are both the formula F = BIL

where F = force in newtons

B = Magnetic flux density in Tesla

I = Current in Amps

L = Length in meters

Explaination- the force on the conductor is due to the two magnetic fields, the stronger the flux density, more force. The more current flowing, the bigger the magnetic field around the conductor and the more force between this field and the magnet. The longer the conductor, the more length this effect is happening over so again the force is increased.

Q3 Magnetic flux density = Flux / area

normally the result would be shown as micro Weber not 0.0375 milli Weber

don't forget to convert square mm to square m by dividing by 1000000(EXP-6)

hope this helps

binky · May 21, 2009

Questions 1 & 2 are both the formula F = BILwhere F = force in newtons

B = Magnetic flux density in Tesla

I = Current in Amps

L = Length in meters

Explaination- the force on the conductor is due to the two magnetic fields, the stronger the flux density, more force. The more current flowing, the bigger the magnetic field around the conductor and the more force between this field and the magnet. The longer the conductor, the more length this effect is happening over so again the force is increased.

Q3 Magnetic flux density = Flux / area

normally the result would be shown as micro Weber not 0.0375 milli Weber

don't forget to convert square mm to square m by dividing by 1000000(EXP-6)

hope this helps

Now has anyone ever applied this formulae whilst working ?:| ? :| ?:|

msb2003 · May 21, 2009

Now has anyone ever applied this formulae whilst working ?:| ? :| ?:|

probably not, but that's the real world......... we're in City and Guilds world here........ totaly different reality!!!!!

searley · May 22, 2009

got number 3 sorted but still can`t get to grips with 1 & 2 do the formulas change around for the two different tests tried everything on the calculator but no joy.

please god tell me i won`t need to remember all this s--te when out in the real world.

MajorOak · May 22, 2009

To get the formula for 1 divide both sides by B*L

So you have

F=B*I*L

dividing both sides by B*L you get

F/B*L = B*I*L / B*L

B and L on the left right side cancel out so you get

F/B*L = I

stick in the numbers

15/0.25*0.5 = 120A

Questing 2 is just plugging in the numbers

F=B*I*L

0.3 * 36 * 0.3 = 3.24N

I think the problem you may have been having is that the formula is in Metres and you were given millimetres?

Hope that helps