NEET 2025

Question

An electron (mass $= 9 	imes 10^{-31}\ 	ext{kg}$ and charge $= 1.6 	imes 10^{-19}\ 	ext{C}$) moving with speed $c/100$ ($c =$ speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 	imes 10^{-4}\ 	ext{T}$ 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 $c = 3 	imes 10^8\ 	ext{ms}^{-1}$) A). $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 	imes 10^4\ 	ext{Vm}^{-1}$ B). $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 	imes 10^4\ 	ext{Vm}^{-1}$ C). $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 	imes 10^2\ 	ext{Vm}^{-1}$ D). $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 	imes 10^2\ 	ext{Vm}^{-1}$

Edvaya · Accepted Answer

The correct answer is $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 	imes 10^2\ 	ext{Vm}^{-1}$. For the electron to travel undeflected, the electric force must balance the magnetic force:
$qE = qvB \Rightarrow E = vB$
$v = c/100 = 3 	imes 10^8/100 = 3 	imes 10^6\ 	ext{ms}^{-1}$
$E = vB = 3 	imes 10^6 	imes 9 	imes 10^{-4} = 27 	imes 10^2\ 	ext{Vm}^{-1} = 2700\ 	ext{Vm}^{-1}$
For the forces to balance: $\vec{E}$ must be **perpendicular to $\vec{B}$** (and also perpendicular to $\vec{v}$), so that $q\vec{E}$ opposes $q\vec{v} 	imes \vec{B}$.

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Physics > Magnetism Induction And Alternating Current > Electromagnetic Waves

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