通俗易懂一些，涉及到微积分的部分，详细展开一下。做到每个步骤都有据可查。---**Problem:**
18. (16分) 如图所示, 平行轨道的间距为L, 轨道平面与水平面夹角为α, 二者的交线与轨道垂直, 以轨道上O点为坐标原点, 沿轨道向下为x轴正方向建立坐标系。轨道之间存在区域 I、II, 区域 I (−2L≤x<−L) 内充满磁感应强度大小为B、方向竖直向上的匀强磁场; 区域 II (x≥0) 内充满方向垂直轨道平面向上的磁场, 磁感应强度大小B₁=k₁t+k₂x, k₁和k₂均为大于零的常量, 该磁场可视为由随时间t均匀增加的匀强磁场和随x坐标均匀增加的磁场叠加而成。将质量为m、边长为L、电阻为R的匀质正方形闭合金属框epqf放置在轨道上, pq边与轨道垂直, 由静止释放。已知轨道绝缘、光滑、足够长且不可移动, 磁场上、下边界均与x轴垂直, 整个过程金属框不发生形变, 重力加速度大小为g, 不计自感。
**(1)** 若金属框从开始进入到完全离开区域 I 的过程中匀速运动, 求金属框匀速运动的速率v和释放时pq边与区域 I 上边界的距离s;
**(2)** 金属框沿轨道下滑, 当ef边刚进入区域 II 时开始计时 (t=0) , 此时金属框的速率为v₀, 若k₁=mgrsinak₂L⁴, 求从开始计时到金属框达到平衡状态的过程中, ef边移动的距离d。
**Chart/Diagram Description:**
* **Type:** Perspective view drawing showing parallel tracks on an inclined plane with magnetic field regions.
* **Main Elements:**
* **Parallel Tracks:** Two parallel lines representing tracks, separated by distance L. They are on an inclined plane.
* **Inclined Plane:** A plane containing the tracks, making an angle α with the horizontal plane (indicated by a dashed line).
* **Coordinate System:** Origin O is on the tracks. X-axis points downwards along the tracks.
* **Regions I and II:** Shaded areas representing magnetic field regions.
* Region I: Located between x = -2L and x = -L. Shaded light grey.
* Region II: Located for x ≥ 0. Shaded grey.
* **Magnetic Field Direction:**
* Region I: Arrow labeled 'B' pointing vertically upwards.
* Region II: Arrow labeled 'B' pointing upwards, perpendicular to the inclined plane.
* **Metal Frame:** A square loop labeled 'epqf' with side length L. Side 'pq' is perpendicular to the tracks.
* **Labels:** Distances 's' and 'L' are labeled along the tracks, indicating the size and position of the regions and the frame. '区域 I' (Region I) and '区域 II' (Region II) are labeled within the respective areas. Angle 'α' is labeled between the inclined plane and the horizontal dashed line.
* **Relative Position and Direction:** The frame is placed on the tracks. Region I is to the left of O (negative x). Region II starts at or to the right of O (positive x). The frame moves downwards along the tracks.
**Other Relevant Text:**
* Problem number: 18.
* Points allocated: (16分)
* Constants: k₁ and k₂ are greater than zero.
* Frame properties: Mass m, side length L, resistance R, uniform, square, closed.
* Track properties: Insulated, smooth, long enough, fixed.
* Magnetic field properties: Upper and lower boundaries are perpendicular to the x-axis.
* Process assumption: Frame does not deform.
* Gravity acceleration: Magnitude g.
* Self-induction: Ignored.
* Given condition in Q2: When ef side just enters Region II, time is t=0 and velocity is v₀.
* Given value for k₁ in Q2: k₁=mgrsinak₂L⁴
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这是一个电磁感应问题，涉及一个正方形金属框在倾斜的平行轨道上滑动。金属框会穿过两个不同的磁场区域。区域I包含竖直向上的匀强磁场B，而区域II有一个同时依赖于时间和位置的变磁场。我们需要分析金属框通过这些区域时的运动和感应电流。
电磁感应的核心是法拉第定律，它告诉我们当磁通量发生变化时会产生感应电动势。磁通量等于磁场强度乘以垂直穿过回路的面积。楞次定律则确定了感应电流的方向，总是阻碍引起磁通量变化的原因。
对于第一问，金属框在区域I中匀速运动意味着受力平衡。重力沿轨道的分量mg sin α必须等于安培力BIL。当金属框以速度v切割磁感线时，产生感应电动势BLv，进而产生电流BLv/R。将这个电流代入力平衡方程，我们可以求出匀速运动的速率。
现在我们详细求解第一问。首先建立力平衡方程，重力分量等于安培力。然后利用法拉第定律求出感应电动势和电流。将电流表达式代入力平衡方程，解得匀速运动的速率。最后用能量守恒定律，重力势能转换为动能，求出释放时的距离s。
第二问涉及更复杂的变磁场。区域II的磁场既随时间变化，又随位置变化。这会产生两种感应电动势：一种由磁场时间变化引起，另一种由导体切割磁感线引起。当金属框达到平衡状态时，这两种电动势的代数和为零。利用给定的k₁值，我们可以通过微积分方法求出ef边的移动距离。
For the first part, we analyze the conditions for uniform motion of the metal frame in Region I. When the frame moves at constant velocity v, the gravitational component along the track equals the magnetic force. The induced EMF is BLv, creating current BLv/R, which produces magnetic force B²L²v/R. Setting up force equilibrium gives us the uniform velocity.
Now let's solve the first problem step by step using calculus and physics principles. We start with force balance, apply Faraday's law for induced EMF, use Ohm's law for current, then substitute back into the force equation. Finally, we use energy conservation where gravitational potential energy converts to kinetic energy to find the initial distance s.
For the second problem, we analyze the variable magnetic field in Region II. The field B₁ = k₁t + k₂x creates two types of induced EMF: one from time-varying flux and another from motional EMF. At equilibrium, the total EMF equals zero, leading to a differential equation. This requires calculus to solve for the frame's motion and find the distance traveled by the ef edge.
Now we solve the second problem using calculus. We set up the differential equation from the equilibrium condition, separate variables, and integrate. Using the given relationship for k₁, we can solve for the distance d that the ef edge travels. This involves integrating the differential equation with the specific boundary conditions and the given constraint on k₁.