The energy of an electron is 100eV when it enters an E * B velocity selector that utilizes a cross electric and magnetic field to produce a zero net force on the particle. The magnitude of the magnetic field is 0.085T. The magnitude of the electric field required to produce a straight-line trajectory is A) 5.0 *10⁴ V/m. B) 2.5 * 10⁵ V/m. C) 5.0 * 10⁵ V/m. D) 2.5 * 10⁶ V/m.

Question

Quizplus · Accepted Answer

For a velocity selector, the electric force $F_e = qE$ must balance the magnetic force $F_m = qvB$, where $q$ is the charge of the electron, $E$ is the electric field, $v$ is the velocity of the electron, and $B$ is the magnetic field. Given the energy of the electron is 100 eV, we first convert this energy to joules (since 1 eV = $1.6 \times 10^{-19}$ J), which gives $100 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-17}$ J. The kinetic energy of the electron is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the electron ($9.11 \times 10^{-31}$ kg). Solving for $v$, we get $v = \sqrt{\frac{2KE}{m}}$. Substituting the values, we find $v$, and then use $E = vB$ to find $E$. Given $B = 0.085T$, and using the calculated $v$, we find that $E$ matches the value in choice C, which is $5.0 \times 10^{15} V/m$.

The Energy of an Electron Is 100eV When It Enters