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An electron (mass $9 \times 10^{-31}\text{ kg}$ and charge $1.6 \times 10^{-19}\text{ C}$) moving with speed $c/100$ ($c =$ speed of light) is injected into a magnetic field $B$ of magnitude $9 \times 10^{-4}\text{ T}$ perpendicular to its direction of motion. We wish to apply a uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3 \times 10^8\text{ ms}^{-1}$):
A
$\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^4\text{ V m}^{-1}$
B
$\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^2\text{ V m}^{-1}$
C
$\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^2\text{ V m}^{-1}$
D
$\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^4\text{ V m}^{-1}$
Explanation
For zero net force, $\vec{E}$ must be perpendicular to both $\vec{v}$ and $\vec{B}$, with magnitude $E = vB = (3 \times 10^6)(9 \times 10^{-4}) = 27 \times 10^2\text{ V/m}$.
Detailed Solution
For an undeflected charged particle (velocity selector principle), the Lorentz force must be zero: $q(\vec{E} + \vec{v} \times \vec{B}) = 0 \implies \vec{E} = -(\vec{v} \times \vec{B})$. Thus $\vec{E}$ must be perpendicular to $\vec{B}$ and $\vec{v}$. The velocity is $v = \frac{c}{100} = \frac{3 \times 10^8}{100} = 3 \times 10^6\text{ m/s}$. The required electric field magnitude is $E = vB = (3 \times 10^6\text{ m/s})(9 \times 10^{-4}\text{ T}) = 27 \times 10^2\text{ V m}^{-1}$.