Question

A magnetic field of $8\hat kmT$ exerts a force of $\left( {4.0\hat i + 3.0\hat j} \right) \times {10^{ - 10}}\;N$ a particle having a charge of $5 \times {10^{ - 10}}\;C$ and going in the X-Y plane. Find the velocity of the particle.

Answer 1

Hint: To solve this evaluate the velocity in the X-Y plane we will use the concept of Lorentz force. It is found that magnetic fields exert forces on another moving charge in a field given. The magnetic field exerts on the force a charge $q$ moving with velocity $v$ is called the magnetic Lorentz force.

Formula used:
Magnetic Lorentz force
$ \Rightarrow \vec F = q\left( {\vec v \times \vec B} \right)$
where $\vec F$ is the force, $q$ is a charge, $\vec v$ is velocity, and $\vec B$ is the magnetic field.

Complete Step-by-step solution
Suppose the velocity of the charged particle in the X-Y plane is given as
$ \Rightarrow \vec v = {v_x}\hat i + {v_y}\hat j$
where ${v_x}\hat i$ is the velocity component in X- direction while ${v_y}\hat j$ is the velocity component in Y- direction.
Here a magnetic field is given as
$\vec B = 8\hat kmT$,
where $\hat k$ represents that the magnetic field is acting perpendicular with the direction of force $\vec F$acting on the field given as
$\vec F = \left( {4.0\hat i + 3.0\hat j} \right) \times {10^{ - 10}}\;N$.
The quantity of the charge is given as $q = 5 \times {10^{ - 10}}C$.
Now from the above quantities given we can easily evaluate the relationship between them is given by Lorentz force which is given as a magnetic field exerts on the force a charge $q$ moving with velocity $v$ is called the magnetic Lorentz force
 $ \Rightarrow \vec F = q\left( {\vec v \times \vec B} \right)$ ……….(1)
The above equation $(1)$ can be written in the form of a matrix as
$ \Rightarrow \vec F = q\left[ {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  {{v_x}}&{{v_y}}&{{v_z}} \\
  {{B_x}}&{{B_y}}&{{B_z}}
\end{array}} \right]$Substituting the values of all the quantities
$ \Rightarrow \vec F = q\left[ {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  {{v_x}}&{{v_y}}&0 \\
  0&0&{8m}
\end{array}} \right]$
$ \Rightarrow \left( {4.0\hat i + 3.0\hat j} \right) \times {10^{ - 10}} = q\left[ {\hat i\left( {8m \times {v_y}} \right) - \hat j\left( {{v_x}8m} \right)} \right]$
Comparing the coefficients of $\left( {\hat i} \right)$ on both side
$ \Rightarrow 4.0 \times {10^{ - 10}} = 5 \times {10^{ - 10}} \times 8 \times {10^{ - 3}}{v_y}$
$ \Rightarrow {v_y} = \dfrac{{4.0 \times {{10}^{ - 10}}}}{{5 \times {{10}^{ - 10}} \times 8 \times {{10}^{ - 3}}}}m/s$
$\therefore {v_y} = 100\;m/s$
Hence the velocity of the particle is given by ${v_y} = 100\;m/s$.

Note
Here in this question we have used the concept of vector quantities. Vector quantities evolve the magnitude as well as directions of the quantities also. If the components are perpendicular then we have to use the cross product $\left( {i \times j} \right)$while when the components are in parallel directions then we have to sue the dot product$\left( {i.j} \right)$.