Question

An electron is moving with a velocity $\vec{v}=(2.0\times {{10}^{6}}\,m/s)\hat{i}\,+\,(3.0\times {{10}^{6}}\,m/s)\hat{j}$ in a magnetic field. If the magnetic field present in the region is $\vec{B}=\,(0.030T)\hat{i}\,-\,(0.15T)\hat{j}$. Then, (a) Find the force on the electron; (b) Find the force on a proton moving with the same velocity.

Answer 1

When a charged particle moves in a magnetic field, it experiences a force in a direction perpendicular to the plane formed by velocity vector and magnetic field vector.
Magnitude of this force depends on magnitude of charge, velocity of the charge, magnitude of magnetic field and angle between velocity vector and magnetic field vector.

Formula used:
Force on a charged particle moving in magnetic field, $\vec{F}=q(\vec{v}\times \vec{B})$

Complete step-by-step answer:
(a) A charged particle experiences force when moving through a magnetic field in a direction perpendicular to the plane formed by velocity vector and magnetic field vector. Magnitude of this force depends on magnitude of charge, velocity of the charge, magnitude of magnetic field and angle between velocity vector and magnetic field vector. Mathematically, force action on a charge ($q$) moving with velocity ($\vec{v}$) in a magnetic field of strength $\vec{B}$ is given as
$\vec{F}=q(\vec{v}\times \vec{B})$
In this question, charged particle is electron i.e. $q=e=-1.6\times {{10}^{-19}}C$ which is moving with velocity $\vec{v}=(2.0\times {{10}^{6}}\,m/s)\hat{i}\,+\,(3.0\times {{10}^{6}}\,m/s)\hat{j}$ in a magnetic field $\vec{B}=\,(0.030T)\hat{i}\,-\,(0.15T)\hat{j}$.
Therefore, force acting on this electron is
{{\vec{F}}_{e}}=(-1.6\times {{10}^{-19}})\left\\{ \left[ (2.0\times {{10}^{6}}\,m/s)\hat{i}\,+\,(3.0\times {{10}^{6}}\,m/s)\hat{j} \right]\times \left[ (0.030T)\hat{i}\,-\,(0.15T)\hat{j}) \right] \right\\}
${{\vec{F}}_{e}}=(-1.6\times {{10}^{-19}})\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 2.0\times {{10}^{6}} & 3.0\times {{10}^{6}} & 0 \\\ 0.030 & -0.15 & 0 \\\ \end{matrix} \right|$
Solving this we get,
${{\vec{F}}_{e}}=(-1.6\times {{10}^{-19}})[(0)\hat{i}+(0)\hat{j}+(-0.24\times {{10}^{6}})\hat{k}]$
${{\vec{F}}_{e}}=3.84\times {{10}^{-14}}\hat{k}$
(b) Since the charge of the proton is $-e$. Therefore, force acting on a proton moving with the same velocity in the same magnetic field will be
${{\vec{F}}_{p}}=-e(\vec{v}\times \vec{B})=-{{\vec{F}}_{e}}$
Therefore,
${{\vec{F}}_{p}}=-3.84\times {{10}^{-14}}\hat{k}$

Note: Magnitude of force acting on proton is the same as that of an electron. But the force on the proton will act in direction opposite to that of the electron.
The force acting on charged particles will be zero if direction of motion of the particle lies along the magnetic field.