Force as a function of angle

Problem

Force measurements on $q = 2\;\mu\text{C}$ with $v = 10^5$ m/s in $B = 0{,}1$ T, as the angle $\theta$ between $\vec v$ and $\vec B$ varies. Verify that $F \propto \sin\theta$ and derive the value of $qvB$.

$\theta$ (degrees)	0	30	60	90
$F$ ($10^{-2}$ N)	0	1,0	1,73	2,0

Solution

Theory predicts $F=qvB\sin\theta$. The coefficient is worth: $qvB=2\cdot10^{-6}\cdot10^5\cdot0{,}1=2\cdot10^{-2}\;\text{N}$

Check point by point: $\begin{aligned} \theta=90^\circ:\; & qvB\sin90^\circ=2\cdot10^{-2}\cdot1=2{,}0\cdot10^{-2}\;\text{N}\ \checkmark \\ \theta=60^\circ:\; & qvB\sin60^\circ=2\cdot10^{-2}\cdot0{,}866=1{,}73\cdot10^{-2}\;\text{N}\ \checkmark \\ \theta=30^\circ:\; & qvB\sin30^\circ=2\cdot10^{-2}\cdot0{,}5=1{,}0\cdot10^{-2}\;\text{N}\ \checkmark \\ \theta=0^\circ:\; & qvB\sin0^\circ=0\ \checkmark \end{aligned}$

The data confirm the proportionality $F\propto\sin\theta$.

$\ev{qvB = 2\cdot 10^{-2}\;\text{N};\quad F \propto \sin\theta}$

Links

Topics: Magnetismo Concepts: Forza di Lorentz Skills: Lettura dei grafici