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:
$1: 1$
$1: \sqrt{2}$
$\sqrt{2}: 1$
$1: 2$
The radius of circular path of an electron when subjected to a perpendicular magnetic field is
A negative charge is coming towards the observer. The direction of the magnetic field produced by it will be (as seen by observer)
An electron is moving with a speed of ${10^8}\,m/\sec $ perpendicular to a uniform magnetic field of intensity $B$. Suddenly intensity of the magnetic field is reduced to $B/2$. The radius of the path becomes from the original value of $r$
A proton and an electron both moving with the same velocity $v$ enter into a region of magnetic field directed perpendicular to the velocity of the particles. They will now move in circular orbits such that
A particle having a mass of $10^{- 2} \,kg$ carries a charge of $5 \times 10^{-8}\, C.$ The particle is given an initial horizontal velocity of $10^5\, m/s $ in the presence of electric field $E$ and magnetic field $B.$ To keep the particle moving in a horizontal direction, it is necessary that