When a proton is released from the rest in a room, it starts with an initial acceleration ${a_0}$ ​towards the west. When it is projected towards the north with a speed ${v_0}$​ it moves with an initial acceleration $3{a_0}$​ towards the west. The electric and magnetic fields in the room are
A. $\dfrac{{m{a_0}}}{e}west,\dfrac{{2m{a_0}}}{{e{v_0}}}up$
B. $\dfrac{{m{a_0}}}{e}west,\dfrac{{2m{a_0}}}{{e{v_0}}}down$
C. $\dfrac{{m{a_0}}}{e}east,\dfrac{{3m{a_0}}}{{e{v_0}}}up$
D. $\dfrac{{m{a_0}}}{e}east,\dfrac{{3m{a_0}}}{{e{v_0}}}down$

Hint: Proton is a positively charged particle and when it is at rest it will not be influenced by the magnetic or electric field present in the surrounding area. But when it starts moving with some velocity then the acceleration and the velocity of the particle will be affected by the presence of the electric and magnetic field.

Complete answer:
Let us assume that the electric field present in the room is $E$ and the magnetic field present in the room is $B$.
Now, as we know that the acceleration of the particle when it starts moving from the rest will be given as,
${a_0} = \dfrac{{eE}}{m}$
From this we can write the expression for the electric field as,
$E = \dfrac{{m{a_0}}}{e}$------equation (1)
And the direction will be the same as that of the direction of the acceleration.
So, the electric field is $E = \dfrac{{m{a_0}}}{e}$ in the west direction.
So, this is the value of the electric field. So, now to get the value of the magnetic field we have to consider the other condition which is given in the question.
So, considering the proton is projected towards north with a speed ${v_0}$​ it moves with an initial acceleration $3{a_0}$​ towards west,
$\dfrac{{e{v_0}B + eE}}{m} = 3{a_0}$
$ \Rightarrow e{v_0}B = 3m{a_0} - eE$
$ \Rightarrow e{v_0}B = 3m{a_0} - m{a_0}$, (substituting $eE = m{a_0}$form the equation (1))
$ \Rightarrow e{v_0}B = 2m{a_0}$
$ \Rightarrow B = \dfrac{{2m{a_0}}}{{e{v_0}}}$
So, this is the value of the magnetic field.
To get the direction of the magnetic field we apply Fleming’s left-hand rule and we come to the conclusion that the direction of the magnetic field will be downwards.
So, the magnetic field will be $B = \dfrac{{2m{a_0}}}{{e{v_0}}}$ downwards.

So, the correct answer is “Option B”.

Note:
In general if a charge of any nature is stationary then it will produce only electrostatic field around itself but if it starts moving with some velocity in any direction with respect to the previous stationary position then it will start to produce both electric field and the magnetic field. For example, if a current carrying conductor will produce both electric and magnetic fields.