317. Spira in campo magnetico

[Tarapìa Tapioco]

Una spira circolare, fatta di un sottile filo conduttivo di raggio , resistività e densità di massa e posta lungo un piano orizzontale, cade da grande altezza in un campo magnetico avente una componente verticale che varia con seguendo la legge . La spira, di diametro , durante la caduta mantiene l'orientazione orizzontale e un campo uniforme lungo le direzioni e . Trascurando l'attrito dell'aria, calcolare la velocità finale di caduta della spira.

[Physicsguy51]

Let $\vec{n}$ be the normal to the plane of the ring, then the flux created by the vertical component of the magnetic field is $\Phi=B S=B_0(1+\alpha z) S$, where $S=\frac{\pi d^2}{4}$ - the area of the EMF induction that occurs in the ring,
$$
\mathcal{E}_i=-\Phi^{\prime}(t)=-\left(B_0(1+\alpha z) S\right)^{\prime}=-B_0 S \alpha z^{\prime}(t) \text {.}
$$
The derivative $z^{\prime}(t)=-v_z$ is the projection of the ring velocity onto the $z$ axis. Thus, $\mathcal{E}_i=-B_0 S \alpha\left(-v_z\right)$. Since the ring velocity is directed against the $z$ axis, then $v_z=-v$, where $v$ is the magnitude of the ring velocity and $\mathcal{E}_i=B_0 S \alpha \nu$. An induced current $I=\frac{\mathcal{E}_i}{R}=\frac{B_0 S \alpha v}{R}$ flows through the ring. As a result, an amount of heat $Q=I^2 R \Delta t$ is released in the ring over a period of time $\Delta t$

At a height of $z_1$, the ring has mechanical energy $W_1=m g z_1+\frac{m v^2}{2}$, at a height of $z_2-W_2=m g z_2+\frac{m v^2}{2}$ ( $v= $ const, i.e. the speed of the ring does not change). According to the law of conservation of energy $W_1=W_2+Q \Rightarrow m g z_1=m g z_2+I^2 R \Delta t \Rightarrow m g\left(z_1 z_2\right)=\left(\frac{B_0 S \alpha v}{R}\right )^2 R \Delta t \Rightarrow m g\left(z_1-z_2\right)\left(^*\right) \\ =\frac{\left(B_0 S \alpha v\right)}{R} \Delta t$


The difference $\left(z_1-z_2\right)$ is the distance covered by the ring during uniform motion, therefore $z_1-z_2=v \Delta t$, and equation will take the form:
$$
m g v \Delta t=\frac{\left(B_0 S \alpha v\right)^2}{R} \Delta t=\Rightarrow m g=\frac{\left(B_0 S \alpha\right)^2 v}{R} \Rightarrow v=\frac{m g R}{\left(B_0 S \alpha\right)^2}=\frac{16 m g R}{\left(B_0 \pi d^2 \alpha\right)^2}
$$

where $m$ is mass of loop and $d$ is its diameter.

[Tarapìa Tapioco]

Could you write the process using the LaTex tool on the Forum so that the language of the message is more understandable and readable? Also, I have modified the prompt, so you need to write the final result according to the data provided by the text. Sorry for the mistake.

P.S. Mass is not a datum given by the problem, but must be written as a function of what the exercise delivers.

[Physicsguy51]

Ok.
Let be the normal to the plane of the ring, then the flux created by the vertical component of the magnetic field is , where - the area of the EMF induction that occurs in the ring,

The derivative is the projection of the ring velocity onto the axis. Thus, . Since the ring velocity is directed against the axis, then , where is the magnitude of the ring velocity and . An induced current flows through the ring. As a result, an amount of heat is released in the ring over a period of time

At a height of , the ring has mechanical energy , at a height of ( = const, i.e. the speed of the ring does not change). According to the law of conservation of energy


The difference is the distance covered by the ring during uniform motion, therefore , and equation will take the form:


where is mass of loop and is its diameter. I know that we were given mass density and radius but in my answer you can just substitute for diameter and .

[Physicsguy51]

Hence final answer will be:

[Physicsguy51]

i made a calculation error.

[Tarapìa Tapioco]

The final result is incorrect. Your first result is very very close to the official solution. But, again, you have to write it as a function of only the data provided by the text of the problem. Not only but also is a quantity not shown in the text.

[Physicsguy51]

Ok, this is what I am getting:

[Tarapìa Tapioco]

No, unfortunately, this is not the correct solution. Most likely, the error is due to your confusion between the diameter of the loop which, therefore, has radius and the diameter of the thin conducting wire which, therefore, has radius .

[Physicsguy51]

Oh, it seems like I neglected one of the two.. I assumed the loop to be of radius r...and directly conserved energy