A charge particle is moving in a uniform magnetic field $(2 \hat{i}+3 \hat{j}) T$. If it has an acceleration of $(\alpha \hat{i}-4 \hat{j}) m / s ^{2}$, then the value of $\alpha$ will be.

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

An $\alpha $ particle and a proton travel with same velocity in a magnetic field perpendicular to the direction of their velocities, find the ratio of the radii of their circular path

Two particles $\mathrm{X}$ and $\mathrm{Y}$ having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $\mathrm{X}$ and $\mathrm{Y}$ is :

A particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{ i }(v \neq 0)$ towards a large screen placed in the $Y – Z$ plane at a distance $d.$ If there is a magnetic field $\overrightarrow{ B }= B _{0} \hat{ k },$ the minimum value of $v$ for which the particle will not hit the screen is