$A$ charged particle is moving in a uniform magnetic field in a circular path with radius $R$. When the energy of the particle is doubled,then the new radius will be

$A$ charge of $2.0\,\mu C$ moves with a speed of $3.0 \times 10^6\,m/s$ along the positive $X$-axis. $A$ magnetic field of strength $\vec B = -0.2\,\hat k\,T$ exists in space. What is the magnetic force $(\vec F_m)$ on the charge?

The acceleration of an electron at a moment in a magnetic field $\vec{B} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ is $\vec{a} = x\hat{i} - 2\hat{j} + \hat{k}$. The value of $x$ is

$A$ proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2 \times 10^5 \text{ ms}^{-1}$. When the electric field is switched off,the proton moves along a circular path of radius $2 \text{ cm}$. The magnitude of the electric field is $x \times 10^4 \text{ N/C}$. The value of $x$ is . . . . . . . (Take the mass of the proton $= 1.6 \times 10^{-27} \text{ kg}$ and charge $e = 1.6 \times 10^{-19} \text{ C}$)

An electric charge moving with uniform velocity has

Bohr model is applied to a particle of mass $m$ and charge $q$ moving in a plane under the influence of a transverse magnetic field $B$. The energy of the charged particle in the $n^{\text{th}}$ level will be $[h = \text{Planck's constant}]$