An electron enters a region where magnetic field (\( \vec{B} \)) and electric field (\( \vec{E} \)) are mutually perpendicular, then

it will always move in the direction of \( \vec{B} \)
it will always move in the direction of \( \vec{E} \)
it always possesses circular motion
it can go undeflected also

Correct answer: it can go undeflected also

Solution

An electron can move undeflected if the electric force and magnetic force acting on it are equal and opposite, which occurs when the velocity of the electron satisfies the condition \( v = \frac{E}{B} \).

Related NEET Physics questions

An alternating electric field, of frequency ν, is applied across the dees (radius = R) of a cyclotron that is being used to accelerate protons (mass = m). The operating magnetic field (B) used in the cyclotron and the kinetic energy (K) of the proton beam, produced by it, are given by:
A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α-particle to describe a circle of same radius in the same field?
A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will
A \( 10 \, \text{eV} \) electron is circulating in a plane at right angles to a uniform field at magnetic induction \( 10^{-4} \, \text{Wb/m}^2 \) (\( = 1.0 \, \text{gauss} \)). The orbital radius of the electron is
A uniform magnetic field acts at right angles to the direction of motion of an electron. As a result, the electron moves in a circular path of radius \( 2 \, \text{cm} \). If the speed of electron is doubled, then the radius of the circular path will be
A deuteron of kinetic energy \( 50 \, \text{keV} \) is describing a circular orbit of radius \( 0.5 \, \text{metre} \) in a plane perpendicular to the magnetic field \( \vec{B} \). The kinetic energy of the proton that describes a circular orbit of radius \( 0.5 \, \text{metre} \) in the same plane with the same \( \vec{B} \) is

⚔️ Practice NEET Physics free + battle 1v1 →