$A$ beam of electrons,initially at rest,is accelerated by a potential $V$. This beam experiences a force $F$ in a uniform magnetic field. The accelerating potential is increased to $V^{\prime}$ and the force experienced by the electrons in the same magnetic field becomes $2F$. The ratio $\frac{V}{V^{\prime}}$ is:

$A$ charged particle of mass $m$ and charge $q$ is projected with velocity $v$ into a region of uniform magnetic field $B$ confined between two parallel plates separated by a distance $d$,as shown in the figure. What is the condition for the charged particle not to strike the opposite plate?

$A$ charged particle is released from rest in a region of uniform electric and magnetic fields,which are parallel to each other. The locus of the particle will be

$A$ charged particle moving along a straight line path enters a uniform magnetic field of $4 \ mT$ at right angles to the direction of the magnetic field. If the specific charge of the charged particle is $8 \times 10^7 \ C \ kg^{-1}$,the angular velocity of the particle in the magnetic field is

The region between $y = 0$ and $y = d$ contains a magnetic field $\vec B = B\hat k$. $A$ particle of mass $m$ and charge $q$ enters the region with a velocity $\vec v = v\hat i$. If $d = \frac{mv}{2qB}$,the acceleration of the charged particle at the point of its emergence at the other side is

$A$ particle having a charge of $10.0\,\mu C$ and mass $1\,\mu g$ moves in a circle of radius $10\,cm$ under the influence of a magnetic field of induction $0.1\,T$. When the particle is at a point $P$, a uniform electric field is switched on so that the particle starts moving along the tangent with a uniform velocity. The electric field is......$V/m$