Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

$\frac{ma_0}{e}$ west, $\frac{ma_0}{ev_0}$ up

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

$\frac{ma_0}{e}$ east, $\frac{2ma_0}{ev_0}$ down

$\frac{ma_0}{e}$ east, $\frac{2ma_0}{ev_0}$ up

Step-by-Step Solution

Analyze Electric Force (Case 1): When released from rest ( $v=0$ ), the magnetic force $F_m = q(\vec{v} \times \vec{B})$ 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 .

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).

Determine Magnetic Field Direction: The magnetic force is given by $\vec{F}_m = e(\vec{v} \times \vec{B})$ . We have $\vec{v}$ towards North and $\vec{F}_m$ towards East. Using the Right-Hand Rule (or vector product): North $\times$ $\vec{B}$ = East.

$(\hat{j} \times \hat{k} = \hat{i})$ . Thus, North $\times$ Up = East. The magnetic field must be directed Upwards .

Calculate Magnitude: $e v_0 B \sin(90^\circ) = 2ma_0$ $B = \frac{2ma_0}{ev_0}$ .
Conclusion: Electric field is $\frac{ma_0}{e}$ East, Magnetic field is $\frac{2ma_0}{ev_0}$ Up.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started