If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field,then the orbit will be or The path executed by a charged particle whose motion is perpendicular to magnetic field is

An electron having kinetic energy of $100 eV$ circulates in a path of radius $10 cm$ in a magnetic field. The magnitude of magnetic field $|B|$ is approximately [Mass of electron $= 0.5 MeV/c^2$,where $c$ is the velocity of light].

An electron enters with a velocity $\vec{v} = v_0 \hat{i}$ into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in a plane parallel to the $xy$-plane. Suggest a configuration of fields $\vec{E}$ and $\vec{B}$ that can lead to this.

$A$ charged particle of $2\,\mu\,C$ accelerated by a potential difference of $100\,V$ enters a region of uniform magnetic field of magnitude $4\,mT$ at a right angle to the direction of the field. The charged particle completes a semicircle of radius $3\,cm$ inside the magnetic field. The mass of the charged particle is $........\times 10^{-18}\,kg$.

Under the influence of a uniform magnetic field,a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

$A$ positively charged $(+q)$ particle of mass $m$ has kinetic energy $K$ and enters vertically downward into a horizontal magnetic field of induction $\vec{B}$. The acceleration of the particle is: