An electron and a proton having same momenta enter perpendicular into a magnetic field then

In a mass spectrometer used for measuring the masses of ions,the ions are initially accelerated by an electric potential $V$ and then made to describe semicircular paths of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant,the ratio $\left(\frac{\text{charge on the ion}}{\text{mass of the ion}}\right)$ will be proportional to

$A$ particle of charge $q$ moves with a velocity $\vec{V} = a \hat{i}$ in a magnetic field $\vec{B} = b \hat{j} + c \hat{k}$,where $a$,$b$,and $c$ are constants. The magnitude of the force experienced by the particle is:

$A$ charged particle with a velocity $2 \times 10^{3} \ ms^{-1}$ passes undeflected through electric and magnetic fields in mutually perpendicular directions. The magnetic field is $1.5 \ T$. The magnitude of the electric field 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: $I = 10 \ A$,$v = 10^5 \ m/s$,$r = 4 \ cm = 0.04 \ m$)

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: