⚛️ Physics

Problem 1 (Easy–Medium): Loop in a Decaying Field

A circular conducting loop of radius $r$ and resistance $R$ is placed in a spatially uniform magnetic field perpendicular to the plane of the loop. The field decays exponentially as $B(t)=B_0{e}^{-t/\tau }$.

(a) Find the induced current $I(t)$.

(b) Find the total charge $Q$ that flows through the loop from $t=0$ to $t=\infty$.

© Explain why $Q$ does not depend on the time constant $\tau$.

Hint: For (b), you can integrate $I(t)$ directly. Or, notice a shortcut: $Q=\int Idt=\frac{1}{R}\int \mathcal{E}dt=\frac{\Delta \Phi }{R}$.

Solution

(a) $\Phi (t)=B(t)\pi r^2=\pi r^2{B}_0{e}^{-\frac{t}{\tau }}$, $I(t)=-\frac{1}{R}\frac{d}{dt}\Phi (t)=\frac{1}{\tau R}\pi r^2{B}_0{e}^{-\frac{t}{\tau }}$
(b) $$Q={\int }_0^{\infty }I(t)dt=\frac{1}{R}(\Phi (0)-\Phi (\infty ))=\frac{1}{R}\pi r^2{B}_0$$
© $Q$ depends only on the total change of magnetic flux.

Problem 2 (Medium–Hard): Falling Chain onto a Scale

A uniform chain of mass $m$ and length $L$ is held vertically with its lower end just touching the pan of a scale. The chain is released from rest and falls freely, piling up on the scale without bouncing.

At the instant when a length $x$ of the chain has landed on the scale, what does the scale read? Express your answer as a function of $x$, and comment on the result.

Hint: The scale must support the weight of the chain already at rest on the pan and provide the force to bring each arriving chain element from speed $v$ to rest. Use $v=\sqrt{2gx}$ for the speed of the chain when length $x$ has landed. For the momentum absorption, consider the impulse $dp=vdm$ delivered in a time interval $dt$.

Solution

Let $\lambda =m/L$ which is the line density. The gravity of the chain already on the scale is $\lambda xg$. Let’s consider the part that just hit the scale $dx$, due to conservation of kinetic energy, $xdx\lambda g=\frac{1}{2}dx\lambda v^2,v=\sqrt{2gx}$, the change in momentum is $dp=\lambda dxv$, so the force that supports the momentum is $F=\lambda v\frac{dx}{dt}=\lambda v^2=2\lambda xg$. The total support force is $3\lambda gx=3\frac{mgx}{L}$. This indicates the instantaneous force is $3$ times the total weight when the chain fully lands the scale.