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- 25-2 일반공학물리(2) 중간고사 문제지-최종본.pdf
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과학

**Processing Exam Problems**

Okay, I've got the PDF loaded and problem extraction is underway. I'm currently working through the process of accurately counting each problem and preparing them for the transformation. My focus is on maintaining the exact original wording and structure. Next, I'll be formatting those problems.


**Initializing Problem Count**

I've set the target problem count and initiated the parallel transformation process. The extracted problems are now being processed in batches of ten. Problem extraction from the document and initial problem definition are complete.


**Formulating Problem Statements**

I'm now fully immersed in constructing the framework for each problem. I've begun to define the key components and variables for each, and the problems are taking shape. For instance, problem one explores calculating electric fields, and problem three involves the definition of a rod with linear charge density. My goal is to capture the essence of each problem clearly and concisely for subsequent transformation. I'm also including all relevant equations and figures.


**Refining Circuit Problems**

I'm presently focusing on problem five, deciphering the RC circuit schematic. The initial assumption about parallel branches was incorrect; it's a more complex configuration. I've re-evaluated the arrangement of the resistors, capacitor, and battery, and I'm updating my initial characterization accordingly. Problem four is progressing well, and I'm refining the description for parallel plates based on the circuit diagram.


**Dissecting Circuit Diagram**

I'm now fully immersed in dissecting the circuit diagram for problem five. Upon closer inspection of the image, I realized the initial series assumption was incorrect. The diagram reveals a more intricate configuration with three branches: a battery and resistor ($R_1$) on the right, a capacitor ($C$) in the middle, and a resistor ($R_2$) on the left. The currents are labeled, and I'm using loop analysis to formulate the equations describing the circuit. Finally, I've re-examined the problem count and confirmed there are five primary problems.


**Transforming Problem Set**

I'm now fully immersed in the parallel transformation of the five exam problems. The images corresponding to problems three, four, and five have been examined, and descriptions for each have been formulated. The focus is on accurately translating the visual information into textual representations. Problem three depicts a rod with variable charge density, problem four shows a setup with parallel plates, and problem five illustrates an RC circuit.


**Examining RC Circuit Details**

I've been meticulously examining the circuit diagram for problem five, using it to refine my understanding. The diagram reveals a battery, a switch, and a more complex arrangement of resistors and a capacitor than initially anticipated. I'm focusing on tracing the currents ($i_1$, $i_2$) and identifying the junctions and loops for the circuit analysis. I have confirmed that the OCR for problem three's equation is accurate. I have also verified the description of problem four's plot. I'm ready to begin.




**Finalizing Problem Sets**

I've executed the tool calls and am ready to deliver the finalized problem sets. However, the `parallel_transform_past_exam_problems` tool's asynchronous nature presents a slight wrinkle. As the main agent in this interaction, waiting for the output seems essential to guarantee the successful presentation of results. I'm now awaiting the tool's final response before continuing.


**Completing My Tasks**

My primary goal has been met. I've successfully called `set_target_problem_count` with the number 5, and then initiated `parallel_transform_past_exam_problems`. The key directive was to delegate the problem transformation, not to perform it myself. I believe my role concludes here, providing a clear handoff to the parallel transformation process. Based on my understanding, I do not need to show the problems, and the system can handle the rest.


**Affirming Problem Handoff**

I have now confirmed that the original problems have been extracted and prepared in the designated XML-hybrid format. Following the instructions, I've successfully transferred the complete set of problems, including their unique identifiers, to the `parallel_transform_past_exam_problems` tool's `originalProblems` parameter. This marks the successful completion of my task of problem delegation.




<problem number="2">
A long, straight, solid conducting wire of radius $R$ carries a non-uniform current density $J$ parallel to the axis of the wire. The magnitude of the current density varies with the distance $r$ from the center of the wire according to $J = J_0 r / R$, where $J_0$ is a constant.

![Cross-section of a long cylindrical conductor with radius R. The current density J flows out of the page and increases linearly with distance r from the center.](cylindrical_conductor_current.png)

(a) Find the total current $I$ carried by the wire in terms of $J_0$ and $R$.
(b) Determine the magnitude of the magnetic field $B$ in the region $r \le R$ and express your result in terms of the total current $I$, $r$, and $R$.

<explanation>
(a) The total current $I$ is the integral of the current density $J$ over the cross-sectional area of the wire:
$I = \int J dA = \int_{0}^{R} \left( J_0 \frac{r}{R} \right) (2\pi r dr)$
$I = \frac{2\pi J_0}{R} \int_{0}^{R} r^2 dr = \frac{2\pi J_0}{R} \left[ \frac{r^3}{3} \right]_{0}^{R} = \frac{2}{3} \pi J_0 R^2$

(b) Using Ampere's Law for a circle of radius $r \le R$:
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{encl}$
$B(2\pi r) = \mu_0 \int_{0}^{r} J(r') 2\pi r' dr'$
$B(2\pi r) = \mu_0 \int_{0}^{r} \left( J_0 \frac{r'}{R} \right) 2\pi r' dr' = \frac{2\pi \mu_0 J_0}{R} \int_{0}^{r} (r')^2 dr' = \frac{2\pi \mu_0 J_0 r^3}{3R}$
$B = \frac{\mu_0 J_0 r^2}{3R}$

From part (a), we have $J_0 = \frac{3I}{2\pi R^2}$. Substituting this into the expression for $B$:
$B = \frac{\mu_0 r^2}{3R} \left( \frac{3I}{2\pi R^2} \right) = \frac{\mu_0 I r^2}{2\pi R^3}$
</explanation>

<answer>
(a) $I = \frac{2}{3} \pi J_0 R^2$
(b) $B = \frac{\mu_0 I r^2}{2\pi R^3}$
</answer>

<source file="25-2 일반공학물리(2) 중간고사 문제지-최종본.pdf" problem="2">Original problem regarding non-uniform charge density in a sphere (Gauss's Law) transformed into non-uniform current density in a cylinder (Ampere's Law) to fit the final exam scope (Chapter 28).</source>
</problem>

---

<problem number="1">
Express the magnetic field $\vec{B}$ and the induced electric field $\vec{E}$ in vector form for the following current distributions in vacuum. Use the permeability of free space $\mu_0$ in the Biot-Savart law and Faraday's law.

(a) A circular wire loop of radius $a$ lies in the $yz$-plane with its center at the origin. It carries a steady current $I$ flowing in the counterclockwise direction when viewed from the positive $x$-axis. Find the magnetic field $\vec{B}$ at a point $P = (x, 0, 0)$ on the axis of the loop.

(b) A long solenoid of radius $R$ has $n$ turns per unit length and carries a time-varying current $I(t) = I_0 \sin(\omega t)$, where $I_0$ and $\omega$ are constants. The axis of the solenoid coincides with the $z$-axis.
(i) Using the principle of superposition (or the result from (a)), find the magnetic field $\vec{B}$ at a point inside the solenoid ($r < R$) near the center, assuming the solenoid is infinitely long.
(ii) Find the induced electric field $\vec{E}$ at a point $P_1$ located at a distance $r$ from the $z$-axis ($r < R$) inside the solenoid. Express your answer in terms of the cylindrical unit vector $\hat{\phi}$ (the tangential direction).

![Diagram showing a circular current loop on the yz-plane and a long solenoid along the z-axis](magnetism_distribution.png)

<explanation>
(a) According to the Biot-Savart Law, the magnetic field contribution $d\vec{B}$ from a current element $Id\vec{l}$ is $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2}$. For a loop in the $yz$-plane, the perpendicular components (in the $y$ and $z$ directions) cancel out due to symmetry. Only the $x$-component survives.
$B_x = \int \frac{\mu_0 I}{4\pi} \frac{dl \sin\theta}{r^2}$, where $r = \sqrt{a^2 + x^2}$ and $\sin\theta = a/r$.
$B_x = \frac{\mu_0 I}{4\pi} \frac{a}{(a^2 + x^2)^{3/2}} \oint dl = \frac{\mu_0 I}{4\pi} \frac{a}{(a^2 + x^2)^{3/2}} (2\pi a) = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$.
Thus, $\vec{B} = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}} \hat{i}$.

(b) (i) For an ideal infinitely long solenoid, the magnetic field is uniform inside and zero outside. Using Ampere's Law $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{encl}$ for a rectangular loop, we find $B = \mu_0 n I$.
Since the current $I(t) = I_0 \sin(\omega t)$ flows along the solenoid, the field points along the $z$-axis: $\vec{B} = \mu_0 n I_0 \sin(\omega t) \hat{k}$.

(ii) Use Faraday's Law in integral form: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$.
For a circular path of radius $r < R$:
$E(2\pi r) = -\frac{d}{dt} \int \vec{B} \cdot d\vec{A} = -\frac{d}{dt} [(\mu_0 n I_0 \sin\omega t) (\pi r^2)]$
$E(2\pi r) = -\mu_0 n I_0 \omega \cos(\omega t) \pi r^2$
$E = -\frac{\mu_0 n I_0 \omega r}{2} \cos(\omega t)$.
In vector form, using the direction given by Lenz's law (represented by $\hat{\phi}$): $\vec{E} = -\frac{\mu_0 n I_0 \omega r}{2} \cos(\omega t) \hat{\phi}$.
</explanation>

<answer>
(a) $\vec{B} = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}} \hat{i}$
(b) (i) $\vec{B} = \mu_0 n I_0 \sin(\omega t) \hat{k}$
(ii) $\vec{E} = -\frac{\mu_0 n I_0 \omega r}{2} \cos(\omega t) \hat{\phi}$
</answer>

<source file="25-2 일반공학물리(2) 중간고사 문제지-최종본.pdf" problem="1">This problem adapts the structure of calculating field distributions from the midterm (Electrostatics) to the final exam scope (Magnetism and Induction), following Chapters 28-30 of University Physics.</source>
</problem>

---

<problem number="5">
Consider the RL circuit shown in the figure below. The circuit consists of a battery with emf $\mathcal{E}$, a switch $SW$, two resistors $R_1$ and $R_2$, and an inductor with inductance $L$. The switch $SW$ is closed at time $t = 0$.

![An RL circuit diagram. On the right vertical branch, there is a battery with emf epsilon and a switch SW. A resistor R1 is located on the top horizontal wire between the right and middle branches. The circuit then splits into two parallel vertical branches: the middle branch contains an inductor L, and the left branch contains a resistor R2.](problem5_varied_image.png)

(a) Write down two simultaneous equations obtained from the application of Kirchhoff's loop rule. 
- The first loop (Loop 1) involves the battery $\mathcal{E}$, the switch $SW$, the resistor $R_1$, and the inductor $L$.
- The second loop (Loop 2) involves the inductor $L$ and the resistor $R_2$.
Express the relations in terms of the current through the inductor $i_L$, the current through the battery branch $i_1$, and the time derivative $di_L/dt$. Assume both loops are traversed in the counter-clockwise direction.

(b) Find the time constant $\tau$ of this circuit, expressed in terms of $R_1$, $R_2$, and $L$.

<explanation>
(a) Let $i_1$ be the current through the battery and $R_1$, $i_L$ be the current through the inductor $L$, and $i_R$ be the current through $R_2$. From the junction rule, $i_1 = i_L + i_R$.
- **Loop 1 (Counter-clockwise)**: Starting from the battery and going through $R_1$ and $L$:
$\mathcal{E} - i_1 R_1 - L \frac{di_L}{dt} = 0$
- **Loop 2 (Counter-clockwise)**: Starting through the inductor and returning through $R_2$:
$L \frac{di_L}{dt} - i_R R_2 = 0 \implies L \frac{di_L}{dt} - (i_1 - i_L) R_2 = 0$

(b) To find the time constant, we need to express the differential equation in terms of $i_L$. 
From Loop 2: $i_1 = \frac{L}{R_2} \frac{di_L}{dt} + i_L$.
Substitute this into the Loop 1 equation:
$\mathcal{E} - \left( \frac{L}{R_2} \frac{di_L}{dt} + i_L \right) R_1 - L \frac{di_L}{dt} = 0$
$\mathcal{E} - i_L R_1 = L \frac{di_L}{dt} \left( 1 + \frac{R_1}{R_2} \right)$
$\mathcal{E} \left( \frac{R_2}{R_1 + R_2} \right) - i_L \left( \frac{R_1 R_2}{R_1 + R_2} \right) = L \frac{di_L}{dt}$
This is in the form $V_{th} - i_L R_{th} = L \frac{di_L}{dt}$, where the equivalent resistance "seen" by the inductor is $R_{th} = \frac{R_1 R_2}{R_1 + R_2}$.
The time constant for an RL circuit is $\tau = \frac{L}{R_{eq}}$.
Thus, $\tau = \frac{L}{R_{th}} = \frac{L (R_1 + R_2)}{R_1 R_2}$.
</explanation>

<answer>
(a) Loop 1: $\mathcal{E} - i_1 R_1 - L \frac{di_L}{dt} = 0$; Loop 2: $L \frac{di_L}{dt} - (i_1 - i_L) R_2 = 0$.
(b) $\tau = \frac{L (R_1 + R_2)}{R_1 R_2}$
</answer>

<source file="25-2 일반공학물리(2) 중간고사 문제지-최종본.pdf" problem="5">The original problem asks for Kirchhoff equations and the time constant of an RC circuit. This variation transforms the capacitor into an inductor to align with Chapter 30 (Inductance) of the final exam range.</source>
</problem>


---


<problem number="4">
Inductor Electronics using Ampere's Law:
Consider a very long solenoid of radius $R$ and $n$ turns per unit length. The solenoid carries a steady current $I_0$ provided by an external power supply, as shown in the figure. The region inside the solenoid is initially filled with air ($\mu \approx \mu_0$).

![Top part: A cross-section of a long solenoid. The magnetic field B points along the z-axis. The radius is R. Bottom part: A graph labeled "< Plot (i) >" showing the magnetic field B on the y-axis versus the radial distance r from the central axis. For 0 < r < R, the value is a constant mu_0*n*I_0. For r > R, the value is 0.](problem4_induction_image.png)

(a) Sketch the following as functions of the radial distance $r$ ($0 \le r \le 2R$):
(i) the magnetic field magnitude $B$ ($T$), (ii) the magnetic energy density $u_B$ ($J/m^3$), and (iii) the magnetic flux $\Phi(r)$ ($Wb$) passing through a circular area of radius $r$ perpendicular to the solenoid axis.
All quantities on the y-axis should be expressed in terms of $R$, $\mu_0$, $n$, and/or $I_0$, as illustrated in Plot (i) above.

(b) Now, a long cylindrical rod of a magnetic material with relative permeability $K_m = 3$ and radius $R/2$ is inserted coaxially into the solenoid (occupying the region $0 \le r \le R/2$).
Repeat the plots for (i), (ii), and (iii) as described above. Assume the current $I_0$ remains constant. You must show all the calculations used to obtain the plots.

(c) Next, instead of the rod in part (b), a hollow cylindrical sleeve of the same magnetic material ($K_m = 3$) is inserted, filling the outer region of the solenoid's interior from $r = R/2$ to $r = R$.
Again, repeat the plots for (i), (ii), and (iii), and show all the corresponding calculations for a constant current $I_0$.

(d) Derive the expressions for the self-inductance per unit length ($L/l$) and the magnetic energy stored per unit length ($U/l$) for each case (i.e., for the configurations in parts (a), (b), and (c)). Which configuration yields the largest stored energy?

<answer>
(a) (i) $B(r) = \mu_0 n I_0$ for $r < R$, $0$ for $r > R$. (ii) $u_B = \frac{B^2}{2\mu_0} = \frac{\mu_0 n^2 I_0^2}{2}$ for $r < R$, $0$ for $r > R$. (iii) $\Phi(r) = \int B dA = \mu_0 n I_0 (\pi r^2)$ for $r < R$, and $\mu_0 n I_0 (\pi R^2)$ for $r \ge R$.

(b) Using Ampere's Law ($\oint \vec{H} \cdot d\vec{l} = I_{encl}$), $H = n I_0$ is constant inside. Thus $B = \mu H = K_m \mu_0 n I_0$.
(i) $B = 3\mu_0 n I_0$ ($0 \le r \le R/2$), $B = \mu_0 n I_0$ ($R/2 < r \le R$), $0$ ($r > R$).
(ii) $u_B = \frac{B^2}{2\mu} = \frac{(3\mu_0 n I_0)^2}{2(3\mu_0)} = 1.5 \mu_0 n^2 I_0^2$ ($r \le R/2$), $u_B = 0.5 \mu_0 n^2 I_0^2$ ($R/2 < r \le R$).
(iii) For $r \le R/2$: $\Phi(r) = 3\mu_0 n I_0 \pi r^2$. For $R/2 < r \le R$: $\Phi(r) = 3\mu_0 n I_0 \pi (R/2)^2 + \mu_0 n I_0 \pi (r^2 - (R/2)^2) = \mu_0 n I_0 \pi (r^2 + \frac{1}{2}R^2)$.

(c) (i) $B = \mu_0 n I_0$ ($0 \le r \le R/2$), $B = 3\mu_0 n I_0$ ($R/2 < r \le R$).
(ii) $u_B = 0.5 \mu_0 n^2 I_0^2$ ($r \le R/2$), $u_B = 1.5 \mu_0 n^2 I_0^2$ ($R/2 < r \le R$).
(iii) For $r \le R/2$: $\Phi(r) = \mu_0 n I_0 \pi r^2$. For $R/2 < r \le R$: $\Phi(r) = \mu_0 n I_0 \pi (R/2)^2 + 3\mu_0 n I_0 \pi (r^2 - (R/2)^2) = \mu_0 n I_0 \pi (3r^2 - \frac{1}{2}R^2)$.

(d) $L/l = n \Phi_{total} / I_0$.
Case (a): $L_a/l = \mu_0 n^2 \pi R^2$, $U_a/l = \frac{1}{2} \mu_0 n^2 I_0^2 \pi R^2$.
Case (b): $\Phi_{total} = 1.5 \mu_0 n I_0 \pi R^2 \implies L_b/l = 1.5 \mu_0 n^2 \pi R^2$, $U_b/l = 0.75 \mu_0 n^2 I_0^2 \pi R^2$.
Case (c): $\Phi_{total} = 2.5 \mu_0 n I_0 \pi R^2 \implies L_c/l = 2.5 \mu_0 n^2 \pi R^2$, $U_c/l = 1.25 \mu_0 n^2 I_0^2 \pi R^2$.
Configuration (c) yields the largest potential energy because the high-permeability material occupies a larger volume (area) of the high-field region.
</answer>

<explanation>
이 문제는 솔레노이드 내부의 자기장과 에너지 밀도, 인덕턴스를 다룹니다.
(a) 공기심 솔레노이드에서 자기장 $B$는 내부에서 균일하고 외부에서 0입니다. 에너지 밀도는 $B^2/2\mu_0$이며, 자속은 면적에 비례하여 증가하다가 $r=R$ 이후 일정해집니다.
(b, c) 자성체($K_m$)가 삽입되어도 앙페르 법칙에 의해 자기 세기 $H = nI$는 일정하게 유지됩니다. 하지만 자기속 밀도 $B = \mu H$는 자성체 내부에서 $K_m$배 증가합니다.
(d) 인덕턴스 $L$은 단위 전류당 총 자속($N\Phi/I$)으로 정의됩니다. 자성체가 차지하는 단면적이 더 넓은 (c)의 경우 자속 변화가 가장 크므로 인덕턴스와 저장된 에너지($U = \frac{1}{2}LI^2$)가 가장 큽니다. 이는 축전기 문제에서 유전체가 삽입되었을 때 정전 용량이 증가하는 것과 물리적으로 대응되는 개념입니다.
</explanation>
</problem>


---

<problem number="3">
A long, straight wire carrying a constant current $I$ is placed along the $z$-axis. A rectangular conducting loop of width $a$ and height $h$ lies in the $xz$-plane. The near side of the loop is at a distance $x$ from the wire, and the loop is moving away from the wire with a constant velocity $v = \frac{dx}{dt}$ in the $+x$-direction, as shown in the figure.

Hint: $\ln(1+x) \simeq x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ for $x \ll 1$.
$(1+x)^{-n} \simeq 1 - nx + \frac{n(n+1)}{2}x^2 - \cdots$ for $x \ll 1$.

![A long vertical wire carrying current I upwards along the z-axis. To its right, in the xz-plane, is a rectangular loop of width a and height h. The distance from the wire to the left edge of the loop is x. The loop is moving to the right with velocity v.](problem3_variant_image.png)

(a) Show that the magnetic flux $\Phi_B$ through the rectangular loop at an instantaneous position $x$ is given by $\Phi_B(x) = \frac{\mu_0 I h}{2\pi} \ln \left( \frac{x + a}{x} \right)$.
(b) Derive an approximate expression for the magnetic flux $\Phi_B(x)$ when the loop is very far from the wire ($x \gg a$). Keep terms up to $(a/x)^2$.
(c) Calculate the induced electromotive force (EMF) $\mathcal{E}$ in the loop using Faraday's Law and the result from part (a). Specify the direction of the induced current (clockwise or counter-clockwise as viewed from the $+y$ direction).
(d) Using the result from part (c), derive an approximate expression for the induced EMF $\mathcal{E}$ in the limit $x \gg a$.
(e) Obtain the induced EMF $\mathcal{E}$ by taking the time derivative of the approximate flux expression found in part (b), and compare it with the result of part (d).

<answer>
(a) The magnetic field from the wire is $B = \frac{\mu_0 I}{2\pi r}$. The flux is $\Phi_B = \int B dA = \int_x^{x+a} \frac{\mu_0 I}{2\pi r} (h dr) = \frac{\mu_0 I h}{2\pi} [\ln r]_x^{x+a} = \frac{\mu_0 I h}{2\pi} \ln(\frac{x+a}{x})$.
(b) $\Phi_B = \frac{\mu_0 I h}{2\pi} \ln(1 + \frac{a}{x}) \approx \frac{\mu_0 I h}{2\pi} \left[ \frac{a}{x} - \frac{a^2}{2x^2} \right]$.
(c) $\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d\Phi_B}{dx} \frac{dx}{dt} = -\frac{\mu_0 I h}{2\pi} \left( \frac{1}{x+a} - \frac{1}{x} \right) v = \frac{\mu_0 I h v a}{2\pi x(x+a)}$. By Lenz's law, as flux (into the page/$-y$ direction) decreases, the induced current must create field in $-y$ direction, so the current is clockwise.
(d) For $x \gg a$, $\mathcal{E} = \frac{\mu_0 I h v a}{2\pi x^2 (1 + a/x)} \approx \frac{\mu_0 I h v a}{2\pi x^2} (1 - \frac{a}{x}) \approx \frac{\mu_0 I h v a}{2\pi x^2}$.
(e) $\mathcal{E} \approx -\frac{d}{dt} \left[ \frac{\mu_0 I h}{2\pi} \frac{a}{x} \right] = -\frac{\mu_0 I h a}{2\pi} (-\frac{1}{x^2} \frac{dx}{dt}) = \frac{\mu_0 I h v a}{2\pi x^2}$. The results are identical.
</answer>

<explanation>
이 문제는 전자기 유도(Faraday's Law)와 비오-사바르 법칙을 결합한 기말고사 범위의 전형적인 문제입니다. 
(a) 무한 직선 도선에 의한 자기장 $B = \mu_0 I / 2\pi r$을 면적에 대해 적분하여 로그 형태의 자속 식을 도출합니다. 
(b) $x \gg a$ 조건에서 $\ln(1+u) \approx u - u^2/2$ 근사를 사용하여 자속을 근사합니다. 
(c) 패러데이 법칙 $\mathcal{E} = -d\Phi/dt$를 연쇄 법칙($\frac{d\Phi}{dx}\frac{dx}{dt}$)을 통해 계산합니다. 렌츠의 법칙에 의해 자속 감소를 방해하는 방향으로 전류가 흐름을 파악합니다. 
(d) 정확한 EMF 식에서 $x \gg a$ 극한을 취하여 $1/x^2$에 비례하는 근사식을 얻습니다. 
(e) 근사된 자속 식을 직접 미분하여 (d)의 결과와 일치함을 확인하며, 이는 물리적 모델의 일관성을 검증하는 과정입니다.
</explanation>

<source file="25-2 일반공ᄒ학물리(2) 중간고사 문제지-최종본.pdf" problem="3">전기적 위치 에너지와 전위 계산 문제를 전자기 유도와 자속 계산 문제로 변형. 로그 함수의 근사 및 극한 거동 분석 구조를 유지함.</source>
</problem>


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