帮我解答这道题---**Extraction Content:**
**Question Number:** 24. (12分)
**Question Stem:**
如图，在直角三角形OPN区域内存在匀强磁场，磁感应强度大小为B、方向垂直于纸面向外。一带正电的粒子从静止开始经电压U加速后，沿平行于x轴的方向射入磁场；一段时间后，该粒子在OP边上某点以垂直于x轴的方向射出。已知O点为坐标原点，N点在y轴上，OP与x轴的夹角为30°，粒子进入磁场的入射点与离开磁场的出射点之间的距离为d，不计重力。求
**Diagram Description:**
* **Type:** Coordinate system with a defined triangular region containing points (magnetic field).
* **Coordinate Axes:** X-axis pointing right, Y-axis pointing up. Origin is labeled as O.
* **Points:** O (origin), N (on the y-axis), P (in the first quadrant).
* **Lines:** ON (part of the y-axis), OP (a line segment starting from O, making an angle of 30° with the positive x-axis), NP (a dashed line segment connecting N and P). These lines form a triangle OPN.
* **Region:** The region inside the triangle OPN is filled with dots, indicating the presence of a magnetic field.
* **Angles:** The angle between the line segment OP and the positive x-axis is labeled as 30°. The angle between the positive x-axis and the positive y-axis is 90° at the origin O.
* **Annotations:** An arrow labeled N is located near the y-axis, pointing horizontally to the right, possibly indicating the direction of entry or location of N on the y-axis boundary.
**Questions Asked:**
(1) 带电粒子的比荷； (Charge-to-mass ratio of the charged particle;)
(2) 带电粒子从射入磁场到运动至x轴的时间。 (Time taken for the charged particle from entering the magnetic field until moving to the x-axis.)
**Other Relevant Text (derived formulas/relations):**
* Particle accelerated by voltage U: $qU = \frac{1}{2}mv_0^2$
* Radius of circular path in magnetic field: $R = \frac{mv_0}{qB}$
* Relation between radius and distance d: $R = \frac{d}{\sqrt{2}}$
* Charge-to-mass ratio: $\frac{q}{m} = \frac{4U}{d^2B^2}$
* Initial velocity after acceleration: $v_0 = \frac{2\sqrt{2}U}{dB}$
* Exit point coordinates: $(x_{out}, y_{out}) = (R, \frac{R}{\sqrt{3}})$
* Time spent in magnetic field: $t_{in} = \frac{\pi d^2B}{8U}$
* Time spent after exiting magnetic field until reaching x-axis: $t_{after} = \frac{d^2B}{4\sqrt{3}U}$
* Total time: $t_{total} = t_{in} + t_{after} = \frac{\pi d^2B}{8U} + \frac{d^2B}{4\sqrt{3}U} = \frac{d^2B}{24U}(3\pi + 2\sqrt{3})$