A particle of charge \[ - 16 \times {10^{ - 18}}\,C\] moving with velocity 10 m/s along the X-axis enters region where a magnetic field of induction B is along Y-axis and field of magnitude \[{10^4}\,V/m\] is along negative Z-axis. If the charged particle continuous moving along the X-axis, the magnitude of B is:

A. \[{10^6}\,Wb/{m^2}\]
B. \[{10^5}\,Wb/{m^2}\]
C. \[{10^3}\,Wb/{m^2}\]
D. \[{10^7}\,Wb/{m^2}\]

Hint: The particle continues to move along the X-axis implies that there is no net force along Z-axis and Y-axis. In that case, the electric force equals the magnetic force on the particle.
Formula used:
\[{F_e} = qE\]
Here, q is the charge of the particle.
\[{F_B} = qvB\]
Here, v is the velocity of the particle.

Complete step by step answer:
We know that the electric force acting on the charged particle placed in the uniform electric field \[E\] is,
\[{F_e} = qE\]
Here, q is the charge of the particle.
Also, the magnetic force on the particle placed in the magnetic field \[B\] is,
\[{F_B} = qvB\]
Here, v is the velocity of the particle.
Since the particle continues to move along the X-axis, the electric force along the negative Z-axis is equal to the magnetic force on the particle along the Y-axis. The net force on the particle is zero except along the X-axis.
Therefore, we can write,
\[qE = qvB\]
\[ \Rightarrow B = \dfrac{E}{v}\]
Substitute \[{10^4}\,V/m\] for E and 10 m/s for v in the above equation.
\[B = \dfrac{{{{10}^4}}}{{10}}\]
\[ \Rightarrow B = {10^3}\,Wb/{m^2}\]

So, the correct answer is “Option C”.

Note:
If the electric force along the negative Z-axis does not equal the magnetic force along the Y-axis, the particle could have moved along the direction whichever the force has maximum value. The unit of magnetic field is Tesla or \[Wb/{m^2}\], therefore, if the magnetic field is given in Tesla, we don’t need to convert it into S.I. unit.