An electron is projected with velocity $v_0$ in a uniform electric field $E$ perpendicular to the field. Again,it is projected with velocity $v_0$ perpendicular to a uniform magnetic field $B$. If $r_1$ is the initial radius of curvature just after entering the electric field and $r_2$ is the initial radius of curvature just after entering the magnetic field,then the ratio $r_1:r_2$ is equal to:

Consider a current-carrying wire as shown in the figure. If the radius of the curved part of the wire is $R$ and the linear parts are assumed to be very long, then the magnetic induction of the field at the point $O$ is

$A$ charge $Q$ is uniformly distributed over the surface of a nonconducting disc of radius $R$. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity $\omega$. As a result of this rotation,a magnetic field of induction $B$ is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity constant and vary the radius of the disc,then the variation of the magnetic induction at the centre of the disc will be represented by which of the following figures?

Two concentric loops $A$ and $B$ of same radius $R = 2 \pi \,cm = 2 \pi \times 10^{-2} \,m$ are placed at right angles to each other. If the currents flowing through $A$ and $B$ are $I_A = 3 \,A$ and $I_B = 4 \,A$ respectively, then the net magnetic field at their common centre is:

$A$ particle of specific charge $(q/m)$ is projected from the origin of coordinates with initial velocity $(u\hat{i} - v\hat{j})$. Uniform electric and magnetic fields exist in the region along the $+y$ direction,of magnitude $E$ and $B$ respectively. The particle will definitely return to the origin once if:

In a Thomson mass spectrograph,the electric field and magnetic field are applied: