A proton with a kinetic energy of $2.0\,eV$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3}\,T$. The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$. The pitch of the helical path taken by the proton is $..........cm$ (Take, mass of proton $=1.6 \times 10^{-27}\,kg$ and Charge on proton $=1.6 \times 10^{-19}\,kg)$
$38$
$41$
$40$
$42$
A particle of charge $q$, mass $m$ enters in a region of magnetic field $B$ with velocity $V_0 \widehat i$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^o$ to its ititial velocity:-
Two ions have equal masses but one is singly ionized and second is doubly ionized. They are projected from the same place in a uniform transverse magnetic field with same velocity then:
$(a)$ Both ions will go along circles of equal radii
$(b)$ The radius of circle described by the single ionized charge is double of radius of circle described by doubly ionized charge
$(c)$ Both circle do not touches to each other
$(d)$ Both circle touches to each other
If a positive ion is moving, away from an observer with same acceleration, then the lines of force of magnetic induction will be
An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform of magnetic field $B$. If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of
Write equation of Lorentz force.