A uniform magnetic field acts at right angles to the direction of motion of electrons. As a result, the electron moves in a circular path of radius $2\, cm$. If the speed of the electrons is doubled, then the radius of the circular path will be.....$cm$

An $\alpha - $ particle travels in a circular path of radius $0.45\, m$ in a magnetic field $B = 1.2\,Wb/{m^2}$ with a speed of $2.6 \times {10^7}\,m/\sec $. The period of revolution of the $\alpha - $ particle is

A magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{j}}$ exists in the region $\mathrm{a} < \mathrm{x} < 2 \mathrm{a}$ and $\vec{B}=-B_0 \hat{j}$, in the region $2 \mathrm{a} < \mathrm{x} < 3 \mathrm{a}$, where $\mathrm{B}_0$ is a positive constant. $\mathrm{A}$ positive point charge moving with a velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\dot{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like,

An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3}\,T$. The radius of path of the circle will approximately be :

When a charged particle enters a uniform magnetic field its kinetic energy

An electron moves with speed $2 \times {10^5}\,m/s$ along the positive $x$-direction in the presence of a magnetic induction $B = \hat i + 4\hat j - 3\hat k$ (in $Tesla$) The magnitude of the force experienced by the electron in Newton's is (charge on the electron =$1.6 \times {10^{ - 19}}C)$