How to calculate magnetic induction just beyond the end of the magnet?

Click For Summary
The discussion centers on calculating the magnetic induction just beyond the end of a bar magnet using the limit expression provided. There is confusion regarding the limit approaching infinity when substituting r with l. It is clarified that the expression was derived under the assumption that r is much greater than l, making it inappropriate for r approaching l. Instead, substituting r with l plus a small increment (epsilon) provides a more accurate approach. The key takeaway is to revisit the original equation for better insights into the behavior of the magnetic induction near the magnet's end.
Lotto
Messages
253
Reaction score
16
Homework Statement
Calculate magnetic induction just beyond the end of a bar magnet on its axis
Relevant Equations
$$B=\frac{\mu}{4\pi}\cdot \frac{4rml}{{\left({r}^2-l^2\right)}^2},$$ where r → l.
Last edited:
Physics news on Phys.org
Lotto said:
Homework Statement:: Calculate magnetic induction just beyond the end of a bar magnet on its axis
Relevant Equations:: $$B=\frac{\mu}{4\pi}\cdot \frac{4rml}{{\left({r}^2-l^2\right)}^2},$$ where r → l.

I know I should use a limit $$B=\lim_{r\to l}{\frac{\mu}{4\pi}\cdot \frac{4rml}{{\left({r}^2-l^2\right)}^2}},$$,but in Wolfram I get a weird solution. https://www.wolframalpha.com/input?i2d=true&i=Limit[Divide[4rl,Power[\(40)Power[r,2]-Power[l,2]\(41),2]],r->l]

What is the solution? It shouldn't be infinity.
According to the expression you quoted, it should be infinity. Note that you are asked to find the magnetic induction just beyond the end of the magnet, not at the end of the magnet. Also, I think your expression was derived in the limit ##r>>l## so you can't do much with it when ##r\approx l##. You need to go back to the original expression from which your equation was derived and see what you get when you substitute ##r=l+\epsilon## with ##\epsilon/l <<1##.
 
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

Replies
10
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
3
Views
306
Replies
1
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K