【Physics】2. Loop in a Decaying Field、Falling Chain onto a Scale

⚛️ Physics

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

A circular conducting loop of radiusrrrand resistanceRRRis placed in a spatially uniform magnetic field perpendicular to the plane of the loop. The field decays exponentially asB(t)=B0e−t/τB(t) = B_0 e^{-t/\tau}B(t)=B0​e−t/τ.

(a) Find the induced currentI(t)I(t)I(t).

(b) Find the total chargeQQQthat flows through the loop fromt=0t = 0t=0tot=∞t = \inftyt=∞.

© Explain whyQQQdoes not depend on the time constantτ\tauτ.

Hint:For (b), you can integrateI(t)I(t)I(t)directly. Or, notice a shortcut:Q=∫I dt=1R∫E dt=ΔΦRQ = \int I\,dt = \frac{1}{R}\int \mathcal{E}\,dt = \frac{\Delta\Phi}{R}Q=∫Idt=R1​∫Edt=RΔΦ​.

Solution

(a)Φ(t)=B(t)πr2=πr2B0e−tτ\Phi(t)=B(t)\pi r^2=\pi r^2B_0e^{-\frac{t}{\tau}}Φ(t)=B(t)πr2=πr2B0​e−τt​,I(t)=−1RddtΦ(t)=1τRπr2B0e−tτI(t)=-\frac{1}{R} \frac{d}{dt} \Phi(t)=\frac{1}{\tau R}\pi r^2B_0 e^{-\frac{t}{\tau}}I(t)=−R1​dtd​Φ(t)=τR1​πr2B0​e−τt​
(b)Q=∫0∞I(t)dt=1R(Φ(0)−Φ(∞))=1Rπr2B0Q=\int_{0}^{\infty}I(t)dt=\frac{1}{R}(\Phi(0)-\Phi(\infty))=\frac{1}{R}\pi r^2B_0Q=∫0∞​I(t)dt=R1​(Φ(0)−Φ(∞))=R1​πr2B0​
©QQQdepends only on the total change of magnetic flux.

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

A uniform chain of massmmmand lengthLLLis 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 lengthxxxof the chain has landed on the scale, what does the scale read? Express your answer as a function ofxxx, and comment on the result.

Hint:The scale must support the weight of the chain already at rest on the panandprovide the force to bring each arriving chain element from speedvvvto rest. Usev=2gxv = \sqrt{2gx}v=2gx​for the speed of the chain when lengthxxxhas landed. For the momentum absorption, consider the impulsedp=v dmdp = v\,dmdp=vdmdelivered in a time intervaldtdtdt.

Solution

Letλ=m/L\lambda =m/Lλ=m/Lwhich is the line density. The gravity of the chain already on the scale isλxg\lambda xgλxg. Let’s consider the part that just hit the scaledxdxdx, due to conservation of kinetic energy,xdxλg=12dxλv2,v=2gxxdx\lambda g=\frac{1}{2}dx\lambda v^2, v=\sqrt{2gx}xdxλg=21​dxλv2,v=2gx​, the change in momentum isdp=λdxvdp=\lambda dx vdp=λdxv, so the force that supports the momentum isF=λvdxdt=λv2=2λxgF=\lambda v\frac{dx}{dt}=\lambda v^2=2\lambda xgF=λvdtdx​=λv2=2λxg. The total support force is3λgx=3mgxL3\lambda gx=3\frac{mgx}{L}3λgx=3Lmgx​. This indicates the instantaneous force is333times the total weight when the chain fully lands the scale.