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 $2.25\times {10}^{-6}$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

(a)Force on proton$=q\overrightarrow{V}\times \overrightarrow{B}$ ($v$=velocity of proton)

$2.25\times {10}^{-6}\hat{j}=1.6\times {10}^{-19}(1.5\times {10}^3\hat{i})\times \overrightarrow{B}$

$\overrightarrow{B}=\frac{2.25\times {10}^{-6}}{1.6\times {10}^{-16}\times 1.5}-\hat{k}$ (negative Z-direction)

$=1.0\times {10}^{10}T-\hat{k}$ (negative Z-direction)

( b) If velocity $V_1=3.2\times {10}^3\frac{m}{s}\widehat{-j}$

$\overrightarrow{B}=1.0\times {10}^{10}T-\hat{k}$

Force on proton $F_p=q{\overrightarrow{V}}_1\times \overrightarrow{B}$

$=1.6\times {10}^{-19}\times 3.2\times {10}^3\times 1.0\times {10}^{10}(-\hat{j}\times -\hat{k})$

$=5.12\times {10}^{6}T\hat{i}$ (positive X -direction)