Right Hand Rule

$\vec{C}=\vec{A}\times \vec{B}$

$\vec{F}=q\vec{V}\times \vec{B}$ $\vec{F}=I\vec{l}\times \vec{B}$

point is out of page +z cross its fleching into page -z

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A charged particle moving in a uniform magnetic field follows a circular trajectory. By Newton’s second law, the magnetic force acting on the particle must be equal to the product of its mass and acceleration:

$$qv\sin \theta =ma.$$

In our scenario, the velocity and field vectors are perpendicular, so $\theta =90^{\circ }$. Also, since the particle moves along a circular path, the acceleration must equal the expression for centripetal acceleration:

$$qvB=\frac{m{v}^2}{r}$$

Solve for vel:

$$v=\frac{qBr}{m}$$

Thus, the speed of a particle can be determined by measuring the radius of its circular path in a known magnetic field, assuming that you also know the charge ($q$) and mass $(m)$ of the particle.-

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The torque a thing experiences depends on the magnetic dipole moment ($\mu$) and it’s orentation w respect to the magnet field. For a current loop its $\mu =IAN$ (current of loop times area of loop times amount of loops times normal of area of loop ). so like $\tau =\mu B\sin \varphi =NIAB\sin \varphi$

imagine a burger:

A particle with a charge of $-1.24\times 10^{-8}$ C is moving with instantaneous velocity $\vec{v}=(4.19\times 10^4\text{ m/s})\hat{i}+(-3.85\times 10^4\text{ m/s})\hat{j}$.

$F=qv\sin \theta$

wait this is so evil how do i do 3d angles 😭 😭

$F=q\vec{v}\times \vec{B}$

$$\begin{array}{rl}F& =q\vec{v}\times \vec{B}\\ & =(-1.24\times 10^{-8})<4.19\times 10^4,-3.85\times 10^4,0>\times \vec{B}\cdot CT\text{m/s}\\ & =(-1.24\times 10^{-8})(10^4)<4.19,-3.85,0>\times \vec{B}\cdot \cancel{(\text{A}\cdot \text{s})\left(\frac{\text{m}}{\text{s}}\right)\left(\frac{\text{kg}}{\text{A}\cdot {\text{s}}^2}\right)}N\end{array}$$

What is the force exerted on this particle by a magnetic field $\vec{B}=(1.70T)\vec{i}$

$$\begin{array}{rl}F& =(-1.24\times 10^{-8})(10^4)<4.19,-3.85,0>\times <1.70,0,0>N\\ & =(-1.24\times 10^{-8})(10^4)\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 4.19& -3.85& 0\\ 1.70& 0& 0\end{array}\right|N\\ & =(-1.24\times 10^{-8})(10^4)\mathbf{i}\left[\text{has a zero}\right]-\mathbf{j}\left[\text{has a zero}\right]+\mathbf{k}[(4.19)(0)-(-3.85)(1.70)]N\\ & =(-1.24\times 10^{-8})(10^4)=\mathbf{i}(0)-\mathbf{j}(0)+\mathbf{k}(0+6.545)N\\ & =(-1.24\times 10^{-8})(10^4)\left[\begin{array}{c}0\\ 0\\ 6.545\end{array}\right]N\end{array}$$

What is the force exerted on this particle by a magnetic field $\vec{B}=(1.70T)\vec{k}$

$$\begin{array}{rl}F& =(-1.24\times 10^{-8})(10^4)<4.19,-3.85,0>\times <0,0,1.70>N\\ & =(-1.24\times 10^{-8})(10^4)\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 4.19& -3.85& 0\\ 0& 0& 1.70\end{array}\right|N\\ & =(-1.24\times 10^{-8})(10^4)\mathbf{i}[(-3.85)(1.70)-(0)(0)]-\mathbf{j}[(4.19)(1.70)-(0)(0)]+\mathbf{k}\left[\text{has a zero}\right]N\\ & =(-1.24\times 10^{-8})(10^4)=\mathbf{i}(-6.545)-\mathbf{j}(7.123)+\mathbf{k}(0)N\\ & =(-1.24\times 10^{-8})(10^4)\left[\begin{array}{c}-6.545\\ -7.123\\ 0\end{array}\right]N\end{array}$$