When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards the east.

Question

When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards the east. When it is projected towards the north with a speed $v_0$, it moves with initial acceleration $3a_0$ towards east. The electric and magnetic fields in the room are:

Accepted Answer

1.  **Analyze Electric Force (Case 1):** When released from rest ($v=0$), the magnetic force $F_m = q(\vec{v} 	imes \vec{B})$ is zero. The only force acting is the electric force $F_e = eE$. 
    *   $F_e = ma_0$ (towards East).
    *   $eE = ma_0 \implies E = \frac{ma_0}{e}$.
    *   Since the proton (positive charge) accelerates East, the Electric Field $\vec{E}$ is directed towards **East** .
2.  **Analyze Magnetic Force (Case 2):** When projected North with speed $v_0$, the total initial acceleration is $3a_0$ towards East. 
    *   Net Force $\vec{F}_{net} = m(3a_0)$ (East).
    *   $\vec{F}_{net} = \vec{F}_e + \vec{F}_m$.
    *   $3ma_0 = ma_0 + F_m \implies F_m = 2ma_0$ (directed towards East).
3.  **Determine Magnetic Field Direction:** The magnetic force is given by $\vec{F}_m = e(\vec{v} 	imes \vec{B})$ .
    *   We have $\vec{v}$ towards North and $\vec{F}_m$ towards East.
    *   Using the Right-Hand Rule (or vector product): North $	imes$ $\vec{B}$ = East.
    *   $(\hat{j} 	imes \hat{k} = \hat{i})$. Thus, North $	imes$ Up = East. The magnetic field must be directed **Upwards** .
4.  **Calculate Magnitude:**
    *   $e v_0 B \sin(90^\circ) = 2ma_0$
    *   $B = \frac{2ma_0}{ev_0}$.
5.  **Conclusion:** Electric field is $\frac{ma_0}{e}$ East, Magnetic field is $\frac{2ma_0}{ev_0}$ Up.

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