(A) For a particle to pass undeflected in a velocity selector,the electric force must balance the magnetic force: $qE = qvB$,which implies $E = vB$.Fi

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(A) For a particle to pass undeflected in a velocity selector,the electric force must balance the magnetic force: $qE = qvB$,which implies $E = vB$.First,calculate the velocity $v$ of the electron using its kinetic energy $K = 728 \, eV$:$K = \frac{1}{2} mv^2$$728 \times 1.6 \times 10^{-19} \, J = \frac{1}{2} \times 9.1 \times 10^{-31} \, kg \times v^2$$v^2 = \frac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$$v^2 = 256 \times 10^{12} \, m^2/s^2$$v = 16 \times 10^6 \, m/s$Now,calculate the electric field $E$:$E = vB = (16 \times 10^6 \, m/s) \times (12 \times 10^{-3} \, T)$$E = 192 \times 10^3 \, V/m = 192 \, kV/m$.

Class 12 Physics Moving Charges and Magnetism Motion of Charged Particle In Magnetic Field

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$A$ velocity selector consists of an electric field $\overrightarrow{E} = E \hat{k}$ and a magnetic field $\overrightarrow{B} = B \hat{j}$ with $B = 12 \, mT$. The value of $E$ required for an electron of energy $728 \, eV$ moving along the positive $x$-axis to pass undeflected is: (Given: mass of electron $= 9.1 \times 10^{-31} \, kg$)

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A
$192 \, kVm^{-1}$
B
$192 \, mVm^{-1}$
C
$9600 \, kVm^{-1}$
D
$16 \, kVm^{-1}$

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

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