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

AI Thread 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
251
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##.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top