I guess this is not a homework check, since i already know the answer and know that my approach is wrong. I want to know, actually, where did i make a mistake:

A hollow cone (like a party hat) has vertex angle 2θ , slant height L, and surface charge density σ . It spins around its symmetry axis with angular frequency ω. What is the magnetic field at the tip?

Now, my point is: The magnetic field of a ring with radius r and current i, at the symmetry axis, is given by $\vec B = \frac{\mu i}{2z} \hat z$

So the field resultant here, is going to be $$\int \frac{\mu \space dI(z)}{2r} \hat z$$

Now, $dI = \frac{dq}{dt} = \frac{\sigma dA}{dt} = \sigma v dz$

$$\vec B = \int \frac{\mu \sigma (v=\omega z tan(\theta)) dz}{2z} \hat z = \frac{\mu \sigma \omega L sin(\theta) }{2} \hat z$$

But the right answer is $$\frac{\mu \sigma \omega L sin^3(\theta) }{2} \hat z$$

I have checked, and i believe that my wrong decision was to consider the field of a infinitesimal area of a cone equals to the field of a ring. But i can't understand why! What is the problem with this approximation?

  • $\begingroup$ Check-my-work questions are also generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. $\endgroup$ Commented Jun 17, 2022 at 7:51

1 Answer 1


The equation $B = \mu I / 2r$ for a circular loop of current only holds for points in the plane of the ring. But the tip of the cone is not in the plane of most of your infinitesimal rings. You will need to use a different equation to find $dB$ in the general case.


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