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

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SUMMARY

The calculation of magnetic induction just beyond the end of a bar magnet involves the limit expression $$B=\lim_{r\to l}{\frac{\mu}{4\pi}\cdot \frac{4rml}{{\left({r}^2-l^2\right)}^2}}$$. Users reported issues with Wolfram Alpha returning unexpected results, suggesting the limit approaches infinity. The correct approach requires substituting $$r=l+\epsilon$$, where $$\epsilon$$ is a small value, to avoid the singularity at $$r=l$$ and accurately determine the magnetic induction.

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  • Understanding of magnetic induction principles
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  • Knowledge of the magnetic permeability constant ($$\mu$$)
  • Experience with Wolfram Alpha for mathematical computations
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  • Learn about the behavior of limits in calculus, particularly near singularities
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Students in physics, particularly those studying electromagnetism, as well as educators and professionals seeking to deepen their understanding of magnetic induction calculations.

Lotto
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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.
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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##.
 

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