Rankers Physics

Force Acting on Moving Charges: Practice Problem & Solution

When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards west. When it is projected towards north with a speed $v_0$ it moves with an initial acceleration $3a_0$ toward west. The electric and magnetic fields in the room are: (2013)
$\frac{ma_0}{e}\text{ east}, \frac{3ma_0}{ev_0}\text{ down}$
$\frac{ma_0}{e}\text{ west}, \frac{2ma_0}{ev_0}\text{ up}$
$\frac{ma_0}{e}\text{ west}, \frac{2ma_0}{ev_0}\text{ down}$
$\frac{ma_0}{e}\text{ east}, \frac{3ma_0}{ev_0}\text{ up}$

Solution Explained:

To solve this problem, we apply the core principles of Force Acting on Moving Charges. Understanding the underlying formula is key to arriving at the correct answer below:

Electric field is $E = \frac{ma_0}{e}$ west from the initial acceleration. With velocity north, the magnetic force accounts for the extra $2a_0$ acceleration west, yielding a downward magnetic field of $B = \frac{2ma_0}{ev_0}$.

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