Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $B = B_0 \hat{k}$.

An electron experiences a force $\vec{F} = (4.0\,\hat{i} + 3.0\,\hat{j}) \times 10^{-13}\,N$ in a uniform magnetic field when its velocity is $\vec{v}_1 = 2.5\,\hat{k} \times 10^7\,m/s$. When the velocity is redirected to $\vec{v}_2 = (1.5\,\hat{i} - 2.0\,\hat{j}) \times 10^7\,m/s$,the magnetic force on the electron is zero. The magnetic field $\vec{B}$ is:

$A$ proton, a deuteron (nucleus of ${ }_1 H^2$) and an $\alpha$-particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is:

Two very long,straight,parallel wires carry steady currents $I$ and $-I$ respectively. The distance between the wires is $d$. At a certain instant of time,a point charge $q$ is at a point equidistant from the two wires,in the plane of the wires. Its instantaneous velocity $v$ is perpendicular to the plane of wires. The magnitude of the force due to the magnetic field acting on the charge at this instant is

$A$ charged particle moves with some initial velocity along the direction of an external magnetic field $B$. Now,if we apply a uniform electric field $E$ perpendicular to the magnetic field,then the trajectory of the charged particle will be

An electron $(e^-)$ is moving parallel to a long current-carrying wire as shown in the figure. Calculate the magnetic force acting on the electron. (Given: $I = 10 \ A$,$v = 10^5 \ m/s$,$r = 4 \ cm = 0.04 \ m$)