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The formula for magnetic flux is generally written as (Formula 1) $$\Phi=AB\cos\theta$$ where $B$ is the magnetic field strength, $A$ is the 'area vector' and $\theta$ is the angle between the NORMAL of the area and the magnetic field.

Is it correct if this formula is re-written as (Formula 2) $$\Phi=AB\sin\alpha$$ where $\alpha$ is the angle between the area and the magnetic field?

This ($\alpha$) is the angle that is generally given in questions and we always need to calculate ($90°-\alpha$) to get the angle $\theta$ needed for Formula 1. Since $\forall x, \cos x=\sin(90°-x)$, can the formaula for magnetic flux just be re-written in terms of this angle $\alpha$?

Unless my calculations are wrong or there is some other special reason I don't know of, would Formula 2 be correct for all applications?

Edit: If it is correct then is there any reason why it isn't used on formula sheets or in textbooks?

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  • $\begingroup$ Are you asking whether $\cos \theta = \sin(90^{\circ} - \theta)$? If not, can you clarify your question. $\endgroup$
    – ProfRob
    Commented Feb 17, 2020 at 8:29
  • $\begingroup$ I know that $\cos\theta=\sin(90°-\theta)$, what I am asking is why the formula for magnetic flux isn't written this way (Formula 2 in my question) because $\alpha$ is always the angle which is given in my questions, and we always need to use ($90°-\alpha$) anyways in calculations. Thanks, I will clarify this in my question. $\endgroup$ Commented Feb 17, 2020 at 9:46

2 Answers 2

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But the definition of a surface is via a normal vector to that surface $d\vec{A}$ at every point on it. The only ambiguity is the 180 degree ambiguity of which way it points (always outward if it is a closed surface, or you can choose otherwise).

The total flux through that surface is $\Phi = \int \vec{B} \cdot d\vec{A}$, which reduces to $BA \cos{\theta}$ only in the case of a constant normal vector and uniform B-field over that area and where the $\cos \theta$ naturally arises from the definition of scalar product.

The version $\Phi = BA\sin \alpha$, where $\alpha = 90^{\circ} -\theta$, is clearly exactly equivalent in that simple case, but how would you express the flux in the more general case of a changing area vector or changing magnetic field vector?

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  • $\begingroup$ Hm. Maybe we should start defining surfaces by two vectors spanning the tangent plane in a sort of 'bivector'. And then B would have to no longer be a vector but some sort of function that takes 'bivectors' and returns a number... $\endgroup$
    – jacob1729
    Commented Feb 17, 2020 at 10:57
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The answer is yes. However, the reason why people did not do it this way is because it is lengthy to calculate the angle $\alpha$.

Think about it, when we have a surface, the mathematical description is always given as the normal vector to that surface. Meanwhile, if you want to find out $\alpha$, you need to first determine the projection of $\vec{B}$ on the surface, then find the angle between the that vector and $\vec{B}$. Hence, this process has to be repeated for different $\vec{B}$, which is really long compared to simple carry out the dot-product which yields the first formula.

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  • $\begingroup$ "it is lengthy to calculate the angle α." ? $\alpha = 90^{\circ} - \theta$. $\endgroup$
    – ProfRob
    Commented Feb 17, 2020 at 8:27
  • $\begingroup$ Consider you are given the plane $x+2y+3z=4$ and $\vec{B} = (5,6,7)$. How are you going to obtain $\alpha$? You have to carry out exactly what I put in my answer. However, if you use the standard formula $\Phi = \vec{B} \cdot \vec{A} = (1,2,3) \cdot (5,6,7) = 5 + 12 + 21 = 38$, it is concise. This dot product is also conveniently defined in terms of $B A \cos \theta$! $\endgroup$ Commented Feb 18, 2020 at 10:07

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