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 ...... $MeV$.

If a charged particle moves in a gravity-free space with uniform velocity,then which of the following is not possible? ($\overrightarrow{E} =$ electric field,$\overrightarrow{B} =$ magnetic field)

$A$ proton moving with a velocity of $2.5 \times 10^7 \, m/s$ enters a magnetic field of intensity $2.5 \, T$ making an angle of $30^o$ with the magnetic field. The force on the proton is:

Uniform magnetic fields of different strengths ($B_1$ and $B_2$),both normal to the plane of the paper,exist as shown in the figure. $A$ charged particle of mass $m$ and charge $q$,at the interface at an instant,moves into the region $2$ with velocity $v$ and returns to the interface. It continues to move into region $1$ and finally reaches the interface. What is the displacement of the particle during this movement along the interface? (Consider the velocity of the particle to be normal to the magnetic field and $B_2 > B_1$)

If an electron enters a magnetic field with its velocity pointing in the same direction as the magnetic field,then

$A$ proton, a deuteron, and an $\alpha$-particle with the same kinetic energy enter a uniform magnetic field at a right angle to the magnetic field. The ratio of the radii of their respective circular paths is