Question

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 $3a_0$ towards west. The electric and magnetic fields in the room are:

• A. $ \frac{ma_0}{e} east, \frac{3ma_0}{ev_0} down$
• B. $ \frac{ma_0}{e} west, \frac{2ma_0}{ev_0} up$
• C. $ \frac{ma_0}{e} west, \frac{2ma_0}{ev_0} down$
• D. $ \frac{ma_0}{e} east, \frac{3ma_0}{ev_0} up$

Answer 1

The Correct Option is C

To determine the electric and magnetic fields in the room, we need to analyze the motion of the proton under the given conditions.

1. Analyzing Initial Condition:
   • When the proton is released from rest, it experiences an initial acceleration $a_0$ towards the west. This indicates the presence of an electric field $\vec{E}$ in the room.
   • The force on the proton due to the electric field is given by $F = eE$, where $e$ is the charge of the proton.
   • Using Newton's second law, $F = ma_0$, we equate it to the electric force: $eE = ma_0$.
   • Thus, the electric field is $E = \frac{ma_0}{e}$ and it is directed towards the west.
2. Analyzing Second Condition:
   • When the proton is projected towards the north with speed $v_0$, it has an initial acceleration of $3a_0$ towards the west.
   • The additional acceleration of $2a_0$ (since the total is $3a_0$ and from the electric field is already $a_0$) must be due to the magnetic field.
   • For a charged particle in a magnetic field, the magnetic force is given by $F = evB$, where $v$ is the velocity and $B$ is the magnetic field.
   • The magnetic force causes an acceleration of $2a_0$, so: $$ ev_0B = m \cdot 2a_0 $$
   • This gives the magnetic field as: $$ B = \frac{2ma_0}{ev_0} $$.
   • Using the right-hand rule and given that the force is directed west while the velocity is north, the magnetic field must be directed downwards.
3. Conclusion:
   • The electric field in the room is $ \frac{ma_0}{e}$ directed west.
   • The magnetic field in the room is $ \frac{2ma_0}{ev_0}$ directed down.
   • Thus, the correct answer is: $ \frac{ma_0}{e} \text{ west, } \frac{2ma_0}{ev_0} \text{ down}$.