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:

A proton and an $\alpha$ -particle, having kinetic energies $K _{ p }$ and $K _{\alpha},$ respectively, enter into $a$ magnetic field at right angles.

The ratio of the radii of trajectory of proton to that of $\alpha$ -particle is $2: 1 .$ The ratio of $K _{ p }: K _{\alpha}$ is :

Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R).$

Assertion $(A)$ : In an uniform magnetic field, speed and energy remains the same for a moving charged particle.

Reason $(R)$ : Moving charged particle experiences magnetic force perpendicular to its direction of motion.

An electron emitted by a heated cathode and accelerated through a potential difference of $ 2.0 \;kV$, enters a region with uniform magnetic field of $0.15\; T$. Determine the trajectory of the electron if the field

$(a)$ is transverse to its initial velocity,

$(b)$ makes an angle of $30^o$ with the initial velocity

A particle of mass $m$ and charge $q$ enters a magnetic field $B$ perpendicularly with a velocity $v$, The radius of the circular path described by it will be

What is the behaviour of perpendicular electric field ${\rm{\vec E}}$ and magnetic field ${\rm{\vec B}}$ ?