The region between $$y = 0$$ and $$y = d$$ contains a magnetic field $$\vec{B} = B\hat{z}$$. A particle of mass m and charge q enters the region with a velocity $$\vec{v} = v\hat{i}$$. If $$d = \frac{mv}{2qB}$$, the acceleration of the charged particle at the point of its emergence at the other side is:

We begin with the Lorentz force law, which gives the magnetic force on a moving charge:

$$\vec F = q(\vec v \times \vec B).$$

Because the magnetic force is always perpendicular to the instantaneous velocity, the particle moves in a circular path of constant speed. The magnitude of the centripetal force equals the magnetic force, so

$$\frac{mv^{2}}{r}=qvB \quad\Longrightarrow\quad r=\frac{mv}{qB}.$$

The problem states that the strip of field extends from $$y=0$$ to $$y=d$$ with

$$d=\frac{mv}{2qB}=\frac{r}{2}.$$

Thus the particle travels only the lower half of the circle’s radius in the $$+y$$-direction while it is inside the field.

To follow the motion geometrically, let us place the centre of the circle at the point $$C(0,r)$$. The particle enters the field at the point $$P_{0}(0,0)$$ with velocity $$\vec v=v\hat i$$. Relative to the centre, the initial radius vector is $$\vec{CP}_{0}= -r\hat j$$, i.e. straight down. We measure an angular coordinate $$\theta$$ from this downward radius, increasing anticlockwise (toward $$+y$$).

Relative to the centre, the position vector is therefore

$$\vec r(\theta)=r\bigl(\sin\theta\,\hat i-\cos\theta\,\hat j\bigr).$$

Differentiating with respect to time gives the velocity direction:

$$\vec v(\theta)=\frac{d\vec r}{d\theta}\frac{d\theta}{dt} =r\dot\theta\bigl(\cos\theta\,\hat i+\sin\theta\,\hat j\bigr) =v\bigl(\cos\theta\,\hat i+\sin\theta\,\hat j\bigr),$$

because $$r\dot\theta=v$$ (the speed is unchanged).

The particle leaves the field when its $$y$$-coordinate equals $$d=r/2$$. The absolute $$y$$-coordinate is

$$y = r - r\cos\theta = r(1-\cos\theta).$$

Setting this equal to $$r/2$$ gives

$$r(1-\cos\theta)=\frac{r}{2}\;\;\Longrightarrow\;\;1-\cos\theta=\frac12 \;\;\Longrightarrow\;\;\cos\theta=\frac12 \;\;\Longrightarrow\;\;\theta=\frac{\pi}{3}\;(60^\circ).$$

Substituting $$\theta=\pi/3$$ into the velocity direction obtains

$$\vec v_{\text{exit}} =v\!\left(\frac12\,\hat i+\frac{\sqrt3}{2}\,\hat j\right).$$

Now we calculate the acceleration at the instant of exit. Using $$\vec a=\dfrac{q}{m}(\vec v\times\vec B)$$ with $$\vec B=B\hat k$$,

$$\vec v_{\text{exit}}\times\vec B =v\!\left(\frac12\,\hat i+\frac{\sqrt3}{2}\,\hat j\right)\!\times B\hat k =vB\!\left[ \frac12(\hat i\times\hat k)+\frac{\sqrt3}{2}(\hat j\times\hat k) \right].$$

Using the right-hand rule $$\hat i\times\hat k=-\hat j$$ and $$\hat j\times\hat k=\hat i$$, we get

$$\vec v_{\text{exit}}\times\vec B =vB\!\left[ \frac12(-\hat j)+\frac{\sqrt3}{2}\hat i \right] =vB\!\left( \frac{\sqrt3}{2}\,\hat i-\frac12\,\hat j \right).$$

Therefore

$$\boxed{\; \displaystyle \vec a_{\text{exit}} =\frac{qvB}{m}\left( \frac{\sqrt3}{2}\,\hat i-\frac12\,\hat j \right) \;} .$$

This vector is not identical to any of the three vectors listed in Options A, B and C, so none of those options is correct.

Hence, the correct answer is Option 4.