# Electromagnetism

Discussion in 'Homework Help' started by user123456, Feb 26, 2014.

1. ### user123456 Thread Starter New Member

Jan 26, 2014
12
0
What is the force of attraction between a north and south pole separated by air? The flux density in the air is 0.84 Tesla and the rectangular poles measure 2cm by 3 cm. (168N)

F = B^2/2u0

0.84^2/2 x 1.257 x 10^-6

Not getting 168Newtons, can anyone please give me a hand here?

2. ### WBahn Moderator

Mar 31, 2012
20,057
5,652
Could you show your work? With units?

If

F = B^2/(2μ0)

What are the units on B?

What are the units on μ0?

What are the units when you divide the first by (twice) the second?

3. ### jjw Member

Dec 24, 2013
194
32
The formula is missing something.
Hint: the area of the magnetic poles has an effect to the force.

4. ### user123456 Thread Starter New Member

Jan 26, 2014
12
0

I multiplied 2 x 3 = 6 cm, still not getting the right figure.

5. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
cm are not SI units, you should be using metres.

further 2cm x 3cm do not multiply to get 6cm.

6. ### user123456 Thread Starter New Member

Jan 26, 2014
12
0
2.66 x 10^-10 if i used meters.

Thanks

7. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,921
600
lol

AREA is squared.

user123456 likes this.
8. ### WBahn Moderator

Mar 31, 2012
20,057
5,652
Well, since the OP can't be troubled to answer these questions, I guess I will.

The units on B are tesla, which is (Ns)/(Cm)

The units on μ0 are H/m or N/(A^2)

Thus, the units of B^2/(2μ0) are [(Ns)^2/(Cm)^2]*[(A^2)/N]

which reduce to N/(m^2)

Thus, a simple dimensional analysis reveals not only that the equation given simply cannot yield the desired answer, but gives a really strong indication of exactly what needs to be done in order to correct it.

But, hey, units are just a troublesome bother that only waste time.

9. ### jjw Member

Dec 24, 2013
194
32
The constant 1.257 x 10^-6 should be in denominator.

10. ### WBahn Moderator

Mar 31, 2012
20,057
5,652
$
F = \frac{B^2 A}{2 \mu_0}
\
F = \frac{(0.84 T)^2 (2 cm \cdot 3 cm)}{2 $$4\pi \cdot 10^{-7} \frac{N}{A^2}$$}
\
F = \frac{ $$0.84 \frac{Ns}{Cm}$$^2 (2 cm \cdot 3 cm)}{2 $$4\pi \cdot 10^{-7} \frac{N}{\(C/s$$^2} \)}
\
F = 1684000 \ $$\frac{ N^2 C^2 s^2 cm^2 }{N C^2 s^2 m^2}$$
\
F = 1684000 \ $$\frac{ N \ cm^2 }{m^2}$$
\
F = 1684000 \ $$\frac{ N \ cm^2 }{m^2}$$ \cdot $$\frac{1 m}{100 cm}$$^2
\
F = 168.4 N
$

See how wasting all that time on tracking units saves a lot of time?

Last edited: Feb 28, 2014
11. ### WBahn Moderator

Mar 31, 2012
20,057
5,652
You are right. That was an editing typo when doing a final fix-up of the LaTeX script.

Thanks for catching it!