As derived from F = qvB sinθ, what is the unit of magnetic force, and why?
Edit: I mistakenly asked about magnetic field before. -- I'm asking the magnetic force.
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\$\begingroup\$ You don't want to memorize, "but [instead to] understand the logic behind it." The answers you've been given don't provide a deeper understanding. (Ask yourself why it is that there might be no magnetic field in one frame of reference, while there is a magnetic field in another. Look into Lagrange's equations of motion, Hamiltonian mechanics, and gauge invariance. Finally, symmetry arguments and reflection space may provide some intuition.) This is probably the wrong place for the logic behind. You might try the physics stackexchange. See if they can help. \$\endgroup\$– jonkCommented Dec 22, 2018 at 3:30
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\$\begingroup\$ @ jonk "reflection space" ?? I don't recall that phrase in the book you recommended. \$\endgroup\$– analogsystemsrfCommented Dec 22, 2018 at 8:57
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\$\begingroup\$ In some reference frames, there is a time difference between various points, resulting in different (orthogonal) forces. We call these (slight, orthogonal) forces the magnetic field. To pretend to understand this in a simple manner, math methods conjure up the magnetic field. TRUE? Note I'm using two words here: forces and fields, as different. \$\endgroup\$– analogsystemsrfCommented Dec 22, 2018 at 9:01
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3 Answers
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It's a force, and that's always measured in Newtons.
B is not A/m, H is measured in A/m. B is measured in \$kg.s^{-2}.A^{-1}\$, base units, or the rather more electrically appropriate volt.seconds per m2, or Webers/m2.
We use sine or cosine depending on what we take to be the reference direction, whether it's along or normal to the conductor. The only difference is 90 degrees.
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while I also don't understand why we use sine but not cosine
Because the actual formula is \$\vec{F}=q\vec{v}\times\vec{B}\$ and the vector cross product is defined so that the magnitidue of the cross product is 0 if the two vectors are parallel, and maximized when the two vectors are orthogonal.
What is the SI unit of magnetic field?
The units are tesla (T).
$$ 1\ {\rm T} = 1 \frac{\rm N \cdot s}{\rm C \cdot m},$$
or, in SI base units
$$ 1\ {\rm T} = 1 \frac{\rm kg}{\rm A \cdot s^2}.$$
answered Dec 21, 2018 at 20:58
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\$\begingroup\$ But if cross-product gives us already the perpendicular vector, then why do we multiply it again with sine ? \$\endgroup\$– VyunCommented Dec 21, 2018 at 21:02
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\$\begingroup\$ The formula you gave is for the magnitude of the force. It isn't a vector equation. To get the magnitude of a cross product, you need to include the sine of the angle between the vectors, \$|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta\$. \$\endgroup\$ Commented Dec 21, 2018 at 21:04
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\$\begingroup\$ I don't understand the reason behind that. \$\endgroup\$– VyunCommented Dec 21, 2018 at 21:11
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\$\begingroup\$ Because we find that this math accurately predicts the forces acting on moving particles subject to magnetic effects. (Note: The field itself is also just a mathematical convenience. The way we define the B field produced by a current or magnetic dipole is also part of "the math" that we need to define to make our predictive model) \$\endgroup\$ Commented Dec 21, 2018 at 21:19
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1\$\begingroup\$ I think I kind of understood. So, it is not something that we conclude logically, but it is because how it (magnitude) can be represented/calculated accurately mathematicaly? I thought the princible was the same as with electrical flux, where E and F are in the same direction, and to find electric field / given area, we multiply E.A.cos (to find the parallel lines and multiply them). I think I need to go over scalar and vectoral multiplications, and I just don't want to memorize, but understand the logic behind it. \$\endgroup\$– VyunCommented Dec 21, 2018 at 22:52
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In SI units, B is measured in teslas (symbol: T) and correspondingly ΦB (magnetic flux) is measured in webers (symbol: Wb) so that a flux density of 1 Wb/m2 is 1 tesla. The SI unit of tesla is equivalent to (newton. second)/(coulomb. metre). - Wikipedia