Question

A group of particles is travelling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at $1.50$ km/s in the $+$x-direction experiences a force of $2.25\times {10}^{-16}$N in the $+$y-direction, and an electron moving at $4.75$ km/s in the $-$z-direction experiences a force of $8.50\times {10}^{-16}$N in the $+$y-direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the $-$y-direction at $3.20$ km/s?

Answer 1

$\left(a\right)$ Let the magnitude field $\overrightarrow{B}$ be $B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$

The magnitude force experienced by proton

$\overrightarrow{F}=q(\overrightarrow{V}\ast \overrightarrow{B})$

$2.25\ast {10}^{-16}\hat{j}=1.6\ast {10}^{-19}\left[\right(1.5\ast {10}^3\hat{i})\ast (B_x\hat{i}+B_y\hat{j}+B_z\hat{k}\left)\right]$

$2.25\ast {10}^{-16}\hat{j}=2.4\ast {10}^{-16}[B_y\hat{k}-B_z\hat{j}]$

By equation $\hat{j},$ we have

$B_z=-0.9375T$

By equation $\hat{k},$ we have

$B_y=0$

$\therefore \overrightarrow{B}=B_x\hat{i}-0.9375\hat{k}$

Now magnetic force experienced by electron

$\overrightarrow{F}=q(\overrightarrow{V}\ast \overrightarrow{B})$

$8.5\ast {10}^{-16}\hat{j}=-1.6\ast {10}^{-19}[4.75\ast {10}^3(-\hat{k})\ast (B_x\hat{i}-0.9375\hat{k}\left)\right]$

$1.118\hat{j}=\left[B_x\right(\widehat{+j}\left)\right]\therefore B_x=1.118T$

Hence $\overrightarrow{B}=1.118\hat{i}-0.9375\hat{k}$

$\therefore \left|\overrightarrow{B}\right|=\sqrt{(1.118{)}^2+(0.9375{)}^2}=1.459T$

$\left(b\right)$ force experienced by electron moving in y-direction

$\overrightarrow{F}=q(\overrightarrow{V}\ast \overrightarrow{B})$

$=-1.6\ast {10}^{-19}[3.2\ast {10}^3(-\hat{j})\ast (1.118\hat{i}-0.9375\hat{k}\left)\right]$

$=5.12\ast {10}^{-16}[-1.118\hat{k}-0.9375\hat{i}]$

$\therefore \overrightarrow{F}=7.47\ast {10}^{-16}$newton