# Help help help cant answer these qs

Discussion in 'Homework Help' started by alex1987, Aug 3, 2009.

1. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
can any help me with this would be much greatfull
1.the cables on a cable try devlop between a magnetic flux denisty of 0.625 teslas calculate the force on a 15 meter of new cable installed on try if it carries a current of 40 amps

2.calculate the inductance of a soleniod cioil of 1200 turns of a copper wire the change in the magnetic field is measured to be an average of 0.05 mwb when the current fluctuates between 2ma and 10ma show all working out

3.a large inductor used to limt current has an inductance of 2.2 henry calculate the voltage induced in the inductor given that the change in inductor current is 6 amperes over a period of 2 milliesconds show all working out

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
1. The force F acting on a conductor of length L carrying a current I in a magnetic field of B Tesla is
F=B*L*I

2. Inductance L [Henry] of a coil of N turns is the total flux linkages N* with the coil per ampere of current I flowing in the coil. Where is the flux in Wb.

L=N*/I

3. Use V=L*(ΔI/Δt) volts

Where L is inductance in Henry, and ΔI is the change in current [amps] over Δt seconds.

3. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
can u show me how the formals would work with my qs still bit stuck

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Sure.

Q1.

B=0.625 Tesla
L=15 meters
I=40 Amps

F=0.625*15*40=375 Newtons

Now you should try the others.

Keep in mind for instance that 1 mWb = .001 Wb.

5. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
why did the answer come out in newtons just trying to full understand it i hope i dont sound to silly

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
It's a reasonable question.

In essence I guess it's about the consistent international system of units that has been adopted for the various derived physical quantities we use to do these calculations. What is a volt really? It has something to to with work (energy) and electrical charge.

I'm not sure how it's done these days, but I have a fuzzy recollection the Ampere was once defined indirectly by measuring (magnetic) force acting on a current carrying conductor. Which is an interesting aspect of your first problem.

I haven't looked at Wikipedia on this but I'll head there now.

7. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
can u show me how to do the other to want to check my answers please

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
May I humbly suggest you give your answers and then I'll give mine.

At some stage you'll need to trust what you do. Now's a good time to start. Even if you are wrong it's no shame - I've been at this stuff for years and I make many mistakes. I'll not think ill of you for trying and making an error - nor I doubt will anyone else who's been doing this for a long time.

9. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
with the 3 qs how do u put the time factor in do u - it or * it

10. ### rspuzio Active Member

Jan 19, 2009
77
0
> Ampere was once defined indirectly by measuring (magnetic) force acting
> on a current carrying conductor.

Yes, the precise definition is that an Ampere is the amount of current
which flowing in two parallel wires separated by a distance d a leads to a
force of 200 nanonewtons on a length d of the wires.

Theoretically, the definition calls for infinitely long wires, but in actual
practice this means that the length should be bigger than the separation
to minimize the relative contribution edge effects (in particular, the wires
are going to have to stop being parallel in order to connect them to the
rest of the circuit). For instance, we might have the wires be 1 m long and
separated by 2 cm. Then we would have a force of 10 millinewtons, which
amounts to roughly the force of a 1 gram weight. This is easily enough
measured and can be measured quite precisely with a carefully constructed
current balance and quality weight set, which is how one would use the
fundamental definition to calibrate an ammeter.

11. ### rspuzio Active Member

Jan 19, 2009
77
0
Correction --- should be 10 micronewtons and
1 milligram, not millinewtons and gram.

12. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
can i please grab the answers for the 3 qs and how the formal is used in relation with qs as i have no idea how to do them still and need them to use them as en example so i can do the rest of my work book thanks

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
See post #4 for Q1

For Q2

The average current = (2mA+10mA)/2=6mA = 6x10^-3 A
The average flux = 0.05mWb = 0.05x10^-3 Wb

L = N x /I= (1200x.05x10^-3)/(6x10^-3) = 10 H

For Q3

V = L x (ΔI/Δt) = 2.2 x (6/(2x10^-3)) = 6.6 kV