JEE Physics 2023 Question with Solution

Given: $\dfrac{q}{m} = 2 \times 10^{11} \, \text{C/kg}$, $v_0 = 3 \times 10^7 \, \text{m/s}$, $\vec{E} = 1.8 \, \hat{j} \, \text{kV/m} = 1.8 \times 10^3 \, \text{V/m}$, and the field region length is $d = 10 \, \text{cm} = 0.1 \, \text{m}$.

Find: Deflection in the $y$-direction.

The electric force acts only along the $y$-direction, so the acceleration is

$$a = \frac{qE}{m}$$

Substituting the given values,

$$a = \left(2 \times 10^{11}\right)\left(1.8 \times 10^3\right) = 3.6 \times 10^{14} \, \text{m/s}^2$$

The time spent inside the electric field is

$$t = \frac{d}{v_0}$$

So,

$$t = \frac{0.1}{3 \times 10^7} = \frac{1}{3 \times 10^8} \, \text{s}$$

Now the deflection in the $y$-direction is

$$y = \frac{1}{2}at^2$$

Substituting,

$$y = \frac{1}{2} \cdot \left(3.6 \times 10^{14}\right) \cdot \left(\frac{1}{3 \times 10^8}\right)^2$$ $$y = \frac{1}{2} \cdot 3.6 \times 10^{14} \cdot \frac{1}{9 \times 10^{16}}$$ $$y = 0.002 \, \text{m}$$

Converting to millimetres,

$$y = 2 \, \text{mm}$$

Therefore, the deflection in the $y$-direction is $2 \, \text{mm}$.

Given: Motion along the $x$-direction with constant speed $v_0$, and electric field along the $+y$-direction.

Principle: The particle has no acceleration in $x$, but it has uniform acceleration in $y$ due to the electric field.

1. Force due to electric field:

$$F = qE$$

2. Acceleration in the $y$-direction:

$$a = \frac{F}{m} = \frac{qE}{m}$$

3. Using the given ratio $q/m$:

$$a = \left(2 \times 10^{11}\right)\left(1.8 \times 10^3\right) = 3.6 \times 10^{14} \, \text{m/s}^2$$

4. Time to cross the field region:

$$t = \frac{0.1}{3 \times 10^7} = \frac{1}{3 \times 10^8} \, \text{s}$$

5. Vertical deflection during this time:

$$y = \frac{1}{2}at^2$$ $$y = \frac{1}{2} \cdot 3.6 \times 10^{14} \cdot \left(\frac{1}{3 \times 10^8}\right)^2$$ $$y = \frac{1.8 \times 10^{14}}{9 \times 10^{16}} = 2 \times 10^{-3} \, \text{m}$$

6. Convert to millimetres:

$$2 \times 10^{-3} \, \text{m} = 2 \, \text{mm}$$

Therefore, the required numerical value is 2.