An $\alpha$-particle of $1 \text{ MeV}$ kinetic energy is moving on a circular path in a uniform magnetic field. The kinetic energy of a proton,to be moved in the same magnetic field on a circular path of double radius,is $..... \text{ MeV}$.

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

An electron projected perpendicular to a uniform magnetic field $B$ moves in a circle. If Bohr's quantization is applicable,then the radius of the electronic orbit in the first excited state is $:$

$A$ proton enters a magnetic field of flux density $1.5 \,Wb \,m^{-2}$ with a velocity of $2 \times 10^7 \,ms^{-1}$ at an angle of $30^{\circ}$ with the field. The force on the proton will be

An $\alpha$-particle of $1 \, MeV$ energy moves on a circular path in a uniform magnetic field. Then,the kinetic energy of a proton in the same magnetic field for a circular path of double radius is ...... $MeV$.

$A$ particle of charge $e$ and mass $m$ moves with a velocity $v$ in a magnetic field $B$ applied perpendicular to the motion of the particle. The radius $r$ of its path in the field is