A positive charge 'q' of mass 'm' is moving along the +x axis. We wish to apply a uniform magnetic field B for time $$\Delta t$$ so that the charge reverses its direction crossing the y axis at a distance d. Then:

A positive charge $$q$$ of mass $$m$$ is moving along the positive x-axis with velocity $$v$$. To reverse its direction and make it cross the y-axis at a distance $$d$$ from the origin, a uniform magnetic field $$B$$ is applied for a time $$\Delta t$$. The force on a moving charge in a magnetic field is given by $$\vec{F} = q(\vec{v} \times \vec{B})$$, which acts as a centripetal force, causing circular motion. Since the charge is positive and initially moving in the $$+x$$ direction, to curve it upwards and cross the y-axis at a positive distance $$d$$, the magnetic field must be directed along the negative z-axis (into the plane).

The magnetic force provides the centripetal force: $$q v B = \frac{m v^2}{r}$$, where $$r$$ is the radius of the circular path. Solving for $$r$$:

$$q v B = \frac{m v^2}{r}$$

$$r = \frac{m v}{q B}$$

Reversing direction requires the charge to traverse a semicircular path. After half a circle, the charge is diametrically opposite its starting point. Assuming it starts at the origin $$(0,0)$$, after a semicircle, it crosses the y-axis at $$(0, d)$$. The distance from the origin to this point is $$d$$, and since the center of the circle is at $$(0, r)$$ (as the initial velocity is horizontal and force is vertical), the y-coordinate after semicircle is $$2r$$. Thus:

$$d = 2r$$

Substituting $$r = \frac{m v}{q B}$$:

$$d = 2 \times \frac{m v}{q B}$$

Solving for $$B$$:

$$d = \frac{2 m v}{q B}$$

$$q B d = 2 m v$$

$$B = \frac{2 m v}{q d}$$

The time $$\Delta t$$ to traverse a semicircle is the distance divided by speed. The arc length of a semicircle is $$\pi r$$, and speed is constant $$v$$, so:

$$\Delta t = \frac{\pi r}{v}$$

Substituting $$r = \frac{d}{2}$$ from $$d = 2r$$:

$$\Delta t = \frac{\pi \times \frac{d}{2}}{v} = \frac{\pi d}{2 v}$$

Thus, $$B = \frac{2 m v}{q d}$$ and $$\Delta t = \frac{\pi d}{2 v}$$. Comparing with the options:

- Option A: $$B = \frac{m v}{q d}$$, $$\Delta t = \frac{\pi d}{v}$$ — Incorrect.
- Option B: $$B = \frac{m v}{2 q d}$$, $$\Delta t = \frac{\pi d}{2 v}$$ — Incorrect.
- Option C: $$B = \frac{2 m v}{q d}$$, $$\Delta t = \frac{\pi d}{2 v}$$ — Correct.
- Option D: $$B = \frac{2 m v}{q d}$$, $$\Delta t = \frac{\pi d}{v}$$ — Incorrect.

Hence, the correct answer is Option C.