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A particle having a mass of $10^{-2}$ kg carries a charge of $5\times10^{-8}$ C. The particle is given an initial horizontal velocity of $10^5\ m\,s^{-1}$ in the presence of electric field $\vec{E}$ and magnetic field $\vec{B}$. To keep the particle moving in a horizontal direction, it is necessary that
(a) $\vec{B}$ should be perpendicular to the direction of velocity and $\vec{E}$ should be along the direction of velocity
(b) Both $\vec{B}$ and $\vec{E}$ should be along the direction of velocity
(c) Both $\vec{B}$ and $\vec{E}$ are mutually perpendicular and perpendicular to the direction of velocity
(d) $\vec{B}$ should be along the direction of velocity and $\vec{E}$ should be perpendicular to the direction of velocity
Which one of the following pairs of statements is possible?
(a) $\vec{B}$ should be perpendicular to the direction of velocity and $\vec{E}$ should be along the direction of velocity
(b) Both $\vec{B}$ and $\vec{E}$ should be along the direction of velocity
(c) Both $\vec{B}$ and $\vec{E}$ are mutually perpendicular and perpendicular to the direction of velocity
(d) $\vec{B}$ should be along the direction of velocity and $\vec{E}$ should be perpendicular to the direction of velocity
Which one of the following pairs of statements is possible?
A
(a) and (c)
B
(c) and (d)
C
(b) and (c)
D
(b) and (d)
Detailed Solution
Force on the particle: $\vec{F} = q\vec{E} + q(\vec{v}\times\vec{B})$. The particle keeps its direction if there is no net force perpendicular to its velocity.
(b): if $\vec{B}$ is along $\vec{v}$, the magnetic force $q(\vec{v}\times\vec{B})$ is zero; with $\vec{E}$ also along $\vec{v}$ the electric force only changes the speed, not the direction. So (b) is possible.
(c): if $\vec{E}$, $\vec{B}$ and $\vec{v}$ are mutually perpendicular, the electric force $q\vec{E}$ and the magnetic force $q(\vec{v}\times\vec{B})$ act along the same line and can be made equal and opposite ($v = \frac{E}{B}$, the velocity selector), so the particle goes undeflected. So (c) is possible.
(a): with $\vec{B}$ perpendicular to $\vec{v}$ there is an unbalanced magnetic force perpendicular to the velocity (the electric force along $\vec{v}$ cannot cancel it), so the path bends.
(d): with $\vec{E}$ perpendicular to $\vec{v}$ and no magnetic force, the electric force deflects the particle.
Hence the possible pair is (b) and (c).
(b): if $\vec{B}$ is along $\vec{v}$, the magnetic force $q(\vec{v}\times\vec{B})$ is zero; with $\vec{E}$ also along $\vec{v}$ the electric force only changes the speed, not the direction. So (b) is possible.
(c): if $\vec{E}$, $\vec{B}$ and $\vec{v}$ are mutually perpendicular, the electric force $q\vec{E}$ and the magnetic force $q(\vec{v}\times\vec{B})$ act along the same line and can be made equal and opposite ($v = \frac{E}{B}$, the velocity selector), so the particle goes undeflected. So (c) is possible.
(a): with $\vec{B}$ perpendicular to $\vec{v}$ there is an unbalanced magnetic force perpendicular to the velocity (the electric force along $\vec{v}$ cannot cancel it), so the path bends.
(d): with $\vec{E}$ perpendicular to $\vec{v}$ and no magnetic force, the electric force deflects the particle.
Hence the possible pair is (b) and (c).