Answer to Question #113025 in Electric Circuits for James

Question

Find the force that acts on a particle with acts on a particle with a charge of 3.2 ×10-19C travelling at a velocity of 5.5 ×103m/s perpendicular to a magnetic field with a flux density of 60mT

Answer 1

A charged particle moving in a magnetic field is affected by the Lorentz force

(1)$\vec F=q[\vec V х\vec B]$ In our problem are given $q=3.2\cdot 10^{-19}C$ , $V=|\vec V|=5.5\cdot10^3m/s$ , $B=|\vec B|=60 mT=60\cdot 10^{-3}T=6.0\cdot 10^{-2}T$ . The vector cross product in (1) can be written as

(2) $[\vec V х\vec B]=\hat n VBsin(\alpha)$ , where $\hat n$ is an unit vector with direction perpendicular to $\vec V$ and $\vec B$ defined by the right hand rule ($\hat k=[\hat iх\hat j])$ ), $\alpha$ is the angle between two vectors $\vec V$ and $\vec B$ . Since we are given that the particle travelling with a velocity perpendicular to a magnetic field $\alpha=90\degree$ and $sin(\alpha)=1$ . If we choose a coordinate system so that $\vec V=V\cdot \hat i$ and $\vec B=B\cdot \hat j$ Lorentz force will be

(3) $\vec F=\hat k \cdot qVB=\hat k \cdot 3.2\cdot 10^{-19}\cdot 5.5 \cdot 10^3\cdot 6.0\cdot 10^{-2}=\hat k \cdot 1.056\cdot 10^{-16}N$

Answer: The force that acts on a particle is $\simeq1.1\cdot 10^{-16}N$