When a proton is released from the rest in a room, it starts with an initial acceleration $a_{0}$ towards north with a speed $v_{0}$ it moves with an initial acceleration $3 a_{0}$ towards west. The electric and magnetic fields in the room are

\frac{m a_{0}}{e}

west,

\frac{2 m a_{0}}{e v_{0}}

\frac{m a_{0}}{e}

west,

\frac{2 m a_{0}}{e v_{0}}

down

\frac{m a_{0}}{e}

west,

\frac{3 m a_{0}}{e v_{0}}

\frac{m a_{0}}{e}

east,

\frac{3 m a_{0}}{e v_{0}}

down

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Solution

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$a_{0} = \frac{e E}{m} \Rightarrow E = \frac{m a_{0}}{e}$

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When a proton is released from the rest in a room, it starts with an initial acceleration a0 towards north with a speed v0 it moves with an initial acceleration 3a0 towards west. The electric and magnetic fields in the room are

a0=eEm⇒E=ma0e

When a proton is released from the rest in a room, it starts with an initial acceleration $a_{0}$ towards north with a speed $v_{0}$ it moves with an initial acceleration $3 a_{0}$ towards west. The electric and magnetic fields in the room are

$a_{0} = \frac{e E}{m} \Rightarrow E = \frac{m a_{0}}{e}$