A particle moving with velocity v having specific charge $(q/m)$ enters a region of magnetic field $B$ having width $d=\frac{{3mv}}{{5qB}}$ at angle $53^o$ to the boundary of magnetic field. Find the angle $\theta$ in the diagram......$^o$
$37$
$60$
$90$
none
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
An electron is accelerated by a potential difference of $12000\, volts$. It then enters a uniform magnetic field of ${10^{ - 3}}\,T$ applied perpendicular to the path of electron. Find the radius of path. Given mass of electron $ = 9 \times {10^{ - 31}}\,kg$ and charge on electron $ = 1.6 \times {10^{ - 19}}\,C$
A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle 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
A uniform magnetic field $B$ and a uniform electric field $E$ act in a common region. An electron is entering this region of space. The correct arrangement for it to escape undeviated is