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 3$a_0$ toward west. The electric and magnetic fields in the room

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

Answer 1

The Correct Option is B

 To determine the electric and magnetic fields in the room based on the given scenarios, we must apply the concept of forces acting on a charged particle in electric and magnetic fields.

1. The force on a proton due to the electric field is given by \(F = eE\), where \(e\) is the charge of the proton and \(E\) is the electric field.
2. The force on a proton moving with velocity \(v\) in a magnetic field \(B\) is given by \(F = evB \sin \theta\), where \(\theta\) is the angle between \(v\) and \(B\). For maximum force, \(\theta = 90^\circ\), hence \(F = evB\).

Let's analyze the scenarios:

• When the proton is released from rest, it experiences an acceleration \(a_0\) towards the west. Thus, the electric field \(E\) must be causing this. Using \(F = ma_0 = eE\), we find: \(E = \frac{ma_0}{e}\). This implies the electric field is directed towards the west.
• When the proton is projected towards the north with speed \(v_0\), it experiences an acceleration \(3a_0\) towards the west. The electric field causes an acceleration \(a_0\), so the additional acceleration \(2a_0\) is due to the magnetic field. The magnetic force is \(ev_0B\), causing an acceleration \(a = \frac{ev_0B}{m}\).
• Equating the magnetic force to the resulting extra acceleration gives: \(B = \frac{2ma_0}{ev_0}\). The right-hand rule indicates that for a force directed towards the west, with velocity towards the north, the magnetic field must be directed downwards.

Hence, the electric and magnetic fields are:

• Electric Field: \(\frac{ma_0}{e}\) west
• Magnetic Field: \(\frac{2ma_0}{ev_0}\) down

Thus, the correct option is:

$\frac{ma_0}{e}$ west $\frac{2ma_0}{ev_0}$ down