A particle moving in a magnetic field increases its velocity, then its radius of the circle
Decreases
Increases
Remains the same
Becomes half
An electron having charge $1.6 \times {10^{ - 19}}\,C$ and mass $9 \times {10^{ - 31}}\,kg$ is moving with $4 \times {10^6}\,m{s^{ - 1}}$ speed in a magnetic field $2 \times {10^{ - 1}}\,tesla$ in a circular orbit. The force acting on electron and the radius of the circular orbit will be
A deutron of kinetic energy $50\, keV$ is describing a circular orbit of radius $0.5$ $metre$ in a plane perpendicular to magnetic field $\overrightarrow B $. The kinetic energy of the proton that describes a circular orbit of radius $0.5$ $metre$ in the same plane with the same $\overrightarrow B $ is........$keV$
A charged particle moves in a magnetic field $\vec B = 10\,\hat i$ with initial velocity $\vec u = 5\hat i + 4\hat j$ The path of the particle will be
An electron is allowed to move with constant velocity along the axis of current carrying straight solenoid.
$A.$ The electron will experience magnetic force along the axis of the solenoid.
$B.$ The electron will not experience magnetic force.
$C.$ The electron will continue to move along the axis of the solenoid.
$D.$ The electron will be accelerated along the axis of the solenoid.
$E.$ The electron will follow parabolic path-inside the solenoid.
Choose the correct answer from the options given below:
A proton and a deutron ( $\mathrm{q}=+\mathrm{e}, m=2.0 \mathrm{u})$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$, moving perpendicular to $\vec{B}$. The ratio of the radius $r_d$ of deutron path to the radius $r_p$ of the proton path is: