Question

A particle of specific charge (charge/mass) starts moving from origin under the action of an electric field $E = {E_0}i$ and magnetic field $B = {B_o}\hat k$. Its velocity at $({x_{0,}}{y_0},0)$ is $(4\hat i - 3\hat j)$. The value of ${x_0}$ is:
(A) $\dfrac{{13a{E_0}}}{{2{B_0}}}$
(B) $\dfrac{{16a{B_0}}}{{{E_0}}}$
(C) $\dfrac{{25}}{{2a{E_0}}}$
(D) $\dfrac{{5a}}{{2{B_0}}}$

Answer 1

Hint
Magnetic forces do not work on a particle irrespective of velocity of the particle because it is perpendicular to the velocity of the particle. By work energy theorem, the work done by an object in a mechanical system with no dissipative forces is equal to the change in energy of the object. So by substituting the values of the work done and the energy, we can calculate the value of ${x_0}$.

Formula used: In this solution we will be using the following formula,
$\Rightarrow W = \Delta E$ where $W$ is the work done on an object, and $\Delta E$ is the change in energy of the object.
Work done by a constant Electric field: $W = qEx$ where $q$ is the charge, $E$ is the electric field and $x$ is the distance traveled by the charge due to the electric field.
Kinetic Energy $KE = \dfrac{1}{2}m{v^2}$ where $m$ is the mass, and $v$ is the velocity of the particle.
In general, $W = \int {F \cdot dr} = \int {Fdr\cos \theta }$ where$W$ is work done,$F$ is force, $r$ is displacement and $\theta$ is the angle between the $F$ and $dr$.

Complete step by step answer
When the charge is moving in the presence of an electric and magnetic field, only the electric field does work on the object because the magnetic field given is always perpendicular to its direction of motion, so no work done by the magnetic field. So work in its general form is given as
$\Rightarrow W = \int {F \cdot dr} = \int {Fdr\cos \theta }$
Hence when calculating work, we will consider only the electric field.
By the work energy theorem, when no dissipative forces are involved, the work done on a body is equal to the change in energy of that body.
Mathematically, we say that
$\Rightarrow W = \Delta E$. $\Delta E$ is all forms of mechanical energy including potential and kinetic energy.
No potential energy was mentioned, we assume change in potential was negligible. Hence
$\Rightarrow W = \Delta KE$ where $\Delta KE$ is equal to the change in kinetic energy of the body.
Work done by an electric field is given as
$\Rightarrow W = qEx$
Hence we can write,
$\Rightarrow qEx = \dfrac{1}{2}m{v^2}$
Making $x$ subject of formula we have
$\Rightarrow x = \dfrac{1}{2}\dfrac{{m{v^2}}}{{qE}}$
At $x = {x_0}$ we have $v = (4\hat i - 3\hat j)$
The magnitude of the velocity is given by
$\Rightarrow v = \sqrt {{4^2} + {3^2}} = 5$
Hence on substituting we get,
$\Rightarrow {x_0} = \dfrac{1}{2}\dfrac{{m\left( {{5^2}} \right)}}{{qE}}$
Since the specific charge is $\dfrac{q}{m}$ which we can write as, $a = \dfrac{q}{m}$
$\Rightarrow {x_0} = \dfrac{1}{2}\dfrac{{25}}{{aE}}$
$\therefore {x_0} = \dfrac{{25}}{{2aE}}$
Hence option (C) is correct.

Note
Alternatively, we can eliminate the magnetic field from mathematical principles as in
Total force on the charge is
$\Rightarrow F = q\vec E + q(\vec v \times \vec B) = m\vec a$ where all variables are in their vector form
Calculating work done we have
$\Rightarrow dW = F \cdot dr = \left[ {q\vec E + q(\vec v \times \vec B)} \right] \cdot dr$
Calculating we have,
$\Rightarrow dW = q\vec E \cdot dr + q(\vec v \times \vec B) \cdot dr$
From mathematical rules, since $dr$and $\vec v$ are in the same direction
$\Rightarrow (\vec v \times \vec B) \cdot dr = 0$
Hence,
$\Rightarrow dW = q\vec E \cdot dr$
$\Rightarrow W = \int {q\vec E \cdot dr}$.