A particle moving in a magnetic field increases its velocity, then its radius of the circle

  • A

    Decreases

  • B

    Increases

  • C

    Remains the same

  • D

    Becomes half

Similar Questions

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........$keV$

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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:

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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:

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