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- 1. An electron is initially at rest, then it is accelerated by a potential difference of 450V. The electron then enters a uniform magnetic field (B = 0.150k T). Calculate the magnitude of the velocity of the electron, and the corresponding radius of path of electron. Solution here, potential difference , V = 450 V let the velocity of electron be v using work energy theorm work done by potential = kinetic energy gained V * q = 0.5 * m * v^2 450 * 1.6 * 10^-19 = 0.5 * 9.1 * 10^-31 * v^2 solving for v v = 1.26 * 10^7 m/s the radius the path of electron , r = m * v /( q * B) r = 9.1 * 10^-31 * 1.26 * 10^7 /( 1.6 * 10^-19 * 0.15) m r = 4.78 *10^-4 m the radius the path of electron is 4.78 *10^-4 m