An electron is projected normally from the surface of a sphere with speed $v_0$ in a uniform magnetic field perpendicular to the plane of the paper such that its strikes symmetrically opposite on the sphere with respect to the $x-$ axis. Radius of the sphere is $'a'$ and the distance of its centre from the wall is $'b'$ . What should be magnetic field such that the charge particle just escapes the wall
$B = \frac{{2bm{v_0}}}{{\left( {{b^2} - {a^2}} \right)e}}$
$B = \frac{{2bm{v_0}}}{{\left( {{a^2} + {b^2}} \right)e}}$
$B = \frac{{m{v_0}}}{{\left( {\sqrt {{b^2} - {a^2}} } \right)e}}$
$B = \frac{{2m{v_0}}}{{\left( {\sqrt {{b^2} - {a^2}} } \right)e}}$
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
An electron moves with a speed of $2 \times 10^5\, m/s$ along the $+ x$ direction in a magnetic field $\vec B = \left( {\hat i - 4\hat j - 3\hat k} \right)\,tesla$. The magnitude of the force (in newton) experienced by the electron is (the charge on electron $= 1.6 \times 10^{-19}\, C$)
An electron (charge $q$ $coulomb$) enters a magnetic field of $H$ $weber/{m^2}$ with a velocity of $v\,m/s$ in the same direction as that of the field the force on the electron is
If a positive ion is moving, away from an observer with same acceleration, then the lines of force of magnetic induction will be
A proton (mass $m$ and charge $+e$) and an $\alpha -$ particle (mass $4m$ and charge $+2e$) are projected with the same kinetic energy at right angles to the uniform magnetic field. Which one of the following statements will be true