A proton released from rest in a room experiences an initial...

Let the directions be represented by unit vectors: East as $\hat{i}$, West as $-\hat{i}$, North as $\hat{j}$, South as $-\hat{j}$, Upward as $\hat{k}$, and Downward as $-\hat{k}$.

When the proton is released from rest, its velocity $\vec{v} = 0$. The only force acting on it is due to the electric field $\vec{E}$.

$\vec{F}_e = e\vec{E} = m\vec{a}$

$e\vec{E} = m(a_0(-\hat{i})) \Rightarrow \vec{E} = \dfrac{ma_0}{e}(-\hat{i})$

Thus, the electric field is $\dfrac{ma_0}{e}$ towards the west.

When the proton is projected towards the north with speed $v_0$, its velocity is $\vec{v} = v_0\hat{j}$. The total force is the vector sum of the electric and magnetic forces.

$\vec{F}_{total} = e\vec{E} + e(\vec{v} \times \vec{B}) = m\vec{a}'$

Given that the new acceleration is $3a_0$ towards the west, we have:

$-ma_0\hat{i} + e(v_0\hat{j} \times \vec{B}) = m(3a_0(-\hat{i}))$

$e(v_0\hat{j} \times \vec{B}) = -2ma_0\hat{i}$

Let the magnetic field be $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$. Substituting this into the cross product:

$ev_0\hat{j} \times (B_x\hat{i} + B_y\hat{j} + B_z\hat{k}) = -2ma_0\hat{i}$

$ev_0(-B_x\hat{k} + B_z\hat{i}) = -2ma_0\hat{i}$

Comparing the components on both sides:

$ev_0B_z = -2ma_0 \Rightarrow B_z = -\dfrac{2ma_0}{ev_0}$

$-ev_0B_x = 0 \Rightarrow B_x = 0$

The component $B_y$ does not contribute to the magnetic force and can be any real number. However, to find the minimum possible magnitude of the magnetic field $|\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2}$, we must set $B_y = 0$.

Therefore, the minimum magnetic field is:

$\vec{B} = -\dfrac{2ma_0}{ev_0}\hat{k}$

Since $\hat{k}$ represents the upward direction, $-\hat{k}$ represents the downward direction. Thus, the minimum magnetic field is $\dfrac{2ma_0}{ev_0}$ downward.