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Question
In a laboratory experiment designed to duplicate Thomson’s determination of $\dfrac{q_e}{m_e}$ , a beam of electrons having a velocity of $6.00\times 10^{7}\textrm{ m/s}$ enters a $5.00\times 10^{-3}\textrm{ T}$ magnetic field. The beam moves perpendicular to the field in a path having a 6.80-cm radius of curvature. Determine $\dfrac{q_e}{m_e}$ from these observations, and compare the result with the known value.
$1.76\times 10^{11}\textrm{ C/kg}$
% error = 0.35%
Solution Video