JEE Physics 2025 Question with Solution

Given: A long straight wire carries current $I$. A particle of mass $m$ and charge $q$ is released from distance $a$ with speed $v_0$ along the wire. It turns back at distance $x$.

Find: The value of $x$ and the correct option.

The solution states that the correct option is D.

The magnetic field due to the wire at distance $r$ is

$$B = \frac{\mu_0 I}{2\pi r}$$

and the magnetic force on the particle is

$$F = qvB = \frac{q\mu_0 I v}{2\pi r}$$

This force is perpendicular to the velocity.

Using the relation presented in the solution,

$$mvr = \text{constant}$$

so initially and at the turning point,

$$mv_0 a = mvx$$

Hence,

$$v = \frac{v_0 a}{x}$$

The provided working then simplifies to the turning-point result

$$x = a \left[1 - \frac{mv_0}{2q\mu_0 I}\right]$$

Therefore, the correct option is D and $x = a \left[1 - \frac{mv_0}{2q\mu_0 I}\right]$.

Given: A current-carrying long straight wire produces magnetic field around it. A charged particle starts at distance $a$ with speed $v_0$ parallel to the current.

Find: The minimum distance $x$ where the particle turns round.

From the solution:

1. Magnetic field due to a long straight wire:

$$B = \frac{\mu_0 I}{2\pi r}$$

2. Magnetic force on the moving charged particle:

$$\vec{F} = q(\vec{v} \times \vec{B})$$

so its magnitude is

$$F = qvB$$

and therefore

$$F = \frac{q\mu_0 I v}{2\pi r}$$

3. The extracted solution uses conservation of angular momentum about the wire axis:

$$mvr = \text{constant}$$

Thus,

$$mv_0 a = mvx$$

which gives

$$v = \frac{v_0 a}{x}$$

4. The working then proceeds with the turning-point analysis and arrives at

$$x = a \left[1 - \frac{mv_0}{2q\mu_0 I}\right]$$

5. The page explicitly concludes that the correct option is D.

Therefore, the value of $x$ is $a \left[1 - \frac{mv_0}{2q\mu_0 I}\right]$.