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Step-by-Step Solution
Step 1: Identify the Forces and Accelerations
A proton (charge = +e, mass = m) experiences two possible forces in the room:
An electric force
$ F_E = e\mathbf{E} $
which produces an acceleration
$ \mathbf{a}_E = \frac{e\mathbf{E}}{m}. $
A magnetic force
$ F_B = e\,(\mathbf{v} \times \mathbf{B}) $
which produces an acceleration
$ \mathbf{a}_B = \frac{e\,(\mathbf{v} \times \mathbf{B})}{m}. $
Step 2: Use the First Condition (Proton Released from Rest)
When the proton is released from rest (velocity = 0), the
magnetic force
$ \mathbf{v} \times \mathbf{B} $
is zero. Hence, the only force acting is the electric force.
This results in an initial acceleration
$ \mathbf{a}_0 $
towards the west. Therefore,
$$
\frac{e\mathbf{E}}{m} = \mathbf{a}_0 \quad \Longrightarrow \quad \mathbf{E} = \frac{m\,\mathbf{a}_0}{e}.
$$
Since the proton accelerates west,
$ \mathbf{E} $
must be directed west.
Step 3: Use the Second Condition (Proton Projected North)
When the proton is given an initial velocity
$ \mathbf{v}_0 $
towards the north, its initial acceleration becomes
$ 3a_0 $
towards the west. Now two contributions exist:
Electric force:
$ \frac{e\mathbf{E}}{m} = \mathbf{a}_0 \text{ (west)}. $
Magnetic force:
$ \mathbf{a}_B = \frac{e}{m} \bigl(\mathbf{v}_0 \times \mathbf{B}\bigr). $
The net acceleration must add up to
$ 3a_0 $
towards the west. Hence:
$$
\mathbf{a}_E + \mathbf{a}_B = 3 \mathbf{a}_0.
$$
But we already know
$ \mathbf{a}_E = \mathbf{a}_0 \text{ (west)}. $
Let
$ \mathbf{a}_B = x\,\mathbf{a}_0 \text{ (west)} $
for some scalar factor x, so that
$$
\mathbf{a}_0 + x\,\mathbf{a}_0 = 3\,\mathbf{a}_0
\quad \Longrightarrow \quad x = 2.
$$
Thus the magnetic contribution to the acceleration is
$ 2 a_0 $
west.
Step 4: Determine the Magnetic Field Vector
From
$ \mathbf{a}_B = \frac{e}{m}(\mathbf{v}_0 \times \mathbf{B}), $
we have:
$$
2a_0 \, \hat{\mathbf{w}} = \frac{e}{m}\Bigl(\mathbf{v}_0 \times \mathbf{B}\Bigr),
$$
where
$ \hat{\mathbf{w}} $
indicates west direction, and
$ \mathbf{v}_0 $
is to the north.
For
$ \mathbf{v}_0 $
north (
$ +\hat{\mathbf{y}} $
) and
$ \mathbf{B} $
downward (
$ -\hat{\mathbf{z}} $
), the cross product
$ \mathbf{v}_0 \times \mathbf{B} $
points west (
$ -\hat{\mathbf{x}}
$ ). Thus:
$$
\mathbf{v}_0 \times \mathbf{B}
= v_0 \, B \, \bigl( \hat{\mathbf{y}} \times -\hat{\mathbf{z}} \bigr)
= v_0\,B\,(-\hat{\mathbf{x}}).
$$
Therefore,
$$
2\,a_0 = \frac{e}{m} \Bigl( v_0 \,B \Bigr).
$$
Solving for
$ B
$ gives:
$$
B = \frac{2\,a_0\,m}{e\,v_0}.
$$
Since the cross product needs to point west, we confirm
$ \mathbf{B} $
must be directed downward.
Step 5: Conclude the Directions and Magnitudes
From the above steps:
$ \mathbf{E} = \frac{m\,a_0}{e} $ (directed west),
$ \mathbf{B} = \frac{2\,m\,a_0}{e\,v_0} $ (directed downward).
This matches the correct answer:
$ \mathbf{E} = \frac{m\,a_0}{e} $ (west),
$ \mathbf{B} = \frac{2\,m\,a_0}{e\,v_0} $ (down)
