How is Magnetic force on a current carrying conductor $Blb$

Question

I was reading an answer about torque acting on a rectangular current carrying loop kept in a uniform magnitude field B. Force acting on each sides is $F_1$, $F_2$, $F_3$, $F_4$. It's written here : Force $F_2$, $F_4$ acting on these sides has same magnitude $F' = Blb \sin(90-\theta) = Blb \cos\theta$. And $F1$ and $F3$, have $F=IlB\sin90° = IlB$.

(where $l$=length; $b$=breadth) I understand $F_1$ and $F_3$. But I don't get how the magnitude of forces of $F_2$, $F_4$ become $Blb \cos\theta$ . Which law says that? Am i missing some knowledge?

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Answer 1

The force that a current carrying wire experiences in a magnetic field is known as the Lorentz force, in differential form: $$ d\mathbf{F} = \int_{wire} I d\mathbf{l}\times\textbf{B}$$.

$d\mathbf{l}$ is in the direction of the current.

Since $I$ is uniform along the wire, the integration is just replaced by $\cdot L$, where $L$ is the length of the wire.

The magnitude of the cross product $| \mathbf{A}\times \mathbf{B} | = |\mathbf{A}|\cdot |\mathbf{B}|\cdot \sin(\phi)$ where $\phi$ is the angle between $\mathbf{A}$ and $\mathbf{B}$.

I am assuming $\theta$ is the angle by which the rectangular loop is deflected with respect to, say, the xy plane.

If sides 2 and and 4 are the ones perpendicular to the xy plane, and $\mathbf{B}$ is in the $z$ direction, then the angle between $d\mathbf{l}$ and $\mathbf{B}$ is $90 - \theta$.