JEE Physics 2023 Question with Solution

Given: Linear charge density $\lambda = 2 \times 10^{-8} \, \text{C/m}$, mass of electron $m = 9 \times 10^{-31} \, \text{kg}$, charge of electron $e = 1.6 \times 10^{-19} \, \text{C}$.

Find: The value of velocity in the form $v = (\_\_) \times 10^6 \, \text{m/s}$.

For an infinite line charge, the electric field at distance $r$ is

$$E = \frac{\lambda}{2\pi \epsilon_0 r}$$

The electrostatic force on the electron is

$$F = eE = \frac{e\lambda}{2\pi \epsilon_0 r}$$

This force provides the centripetal force for circular motion:

$$\frac{mv^2}{r} = \frac{e\lambda}{2\pi \epsilon_0 r}$$

Cancelling $r$ from both sides,

$$v^2 = \frac{e\lambda}{2\pi \epsilon_0 m}$$

Therefore,

$$v = \sqrt{\frac{e\lambda}{2\pi \epsilon_0 m}}$$

Substituting the given values,

$$v = \sqrt{\frac{(1.6 \times 10^{-19})(2 \times 10^{-8})}{2\pi(8.85 \times 10^{-12})(9 \times 10^{-31})}}$$ $$v = \sqrt{\frac{3.2 \times 10^{-27}}{5.0 \times 10^{-41}}}$$ $$v = \sqrt{6.4 \times 10^{13}}$$ $$v \approx 8 \times 10^6 \, \text{m/s}$$

Therefore, the required numerical value is 8.

Given: The electron moves in a circular path around an infinite line charge.

Find: Why the speed can be found without knowing the radius.

The electric field of an infinite line charge varies as $1/r$, so the electrostatic force on the electron also varies as $1/r$. The centripetal force requirement is

$$\frac{mv^2}{r}$$

which also contains the same $1/r$ dependence after equating forces. Hence the radius cancels immediately:

$$\frac{mv^2}{r} = \frac{e\lambda}{2\pi \epsilon_0 r}$$ $$v^2 = \frac{e\lambda}{2\pi \epsilon_0 m}$$

So the speed depends only on $e$, $\lambda$, $\epsilon_0$ and $m$, not on the orbit radius. Substituting values gives $v \approx 8 \times 10^6 \, \text{m/s}$. Therefore, the required numerical value is 8.