Question

A square loop is cut out of a thick sheet of aluminum. It is then placed so that the top portion is in a uniform magnetic field , B and is allowed to fall under gravity (Fig. 7 .20). (In the diagram, shading indicates the field region; points into the page.) If the magnetic field is 1 T (a pretty standard laboratory field), find the terminal velocity of the loop (in m/s ). Find the velocity of the loop as a function of time. How long does it take (in seconds) to reach, say, $90\%$ of the terminal velocity? What would happen if you cut a tiny slit in the ring, breaking the circuit? [Note: The dimensions of the loop cancel out; determine the actual numbers, in the units indicated.]

Short answer

The terminal velocity is$0.0185m/s$ , velocity as function of time,$\left(v_{t}1-e^{-out}\right)$ , and time taken to reach$90\%$ of the terminal velocity is 2.8ms .

Step-by-step solution

write the given data from the question.

The uniform magnetic field is B .

The magnetic field B=1T

The standard value for aluminium.

The mass density of aluminium,$n=2.7\times {10}^{3}kg/m^3$

The resistivity of aluminium,$p=2.8\times {10}^{-8}\Omega -m$

Calculate the terminal velocity of the loop, velocity as function of time, and time taken to reach 90% of terminal voltage.

Let assume the mass of loop is m , resistance of loop is R . length of loop is l and v is the velocity of loop.

Due the motion in the magnetic field of loop, the induced emf of loop is given by,

$\mathit{\epsilon }\mathbf{=}\mathit{B}\mathit{l}\mathit{v}$ …… (1)

The induced emf in terms of current is given by,

$\mathit{\epsilon }\mathbf{=}\mathit{i}\mathit{R}$ …… (2)

Equate the equation (1) and (2),

$$\begin{gathered} BIv=iR \\ i=\frac{BIv}{R} \end{gathered}$$

The force acting on the loop in upward direction is given by,

$F=Bli$

Substitute $\frac{BIv}{R}$for into above equation.

$$\begin{gathered} F=BI\left(\frac{BIv}{R}\right) \\ F=\frac{B^2{I}^{2}v}{R} \end{gathered}$$

The magnetic force that is acting on the loop is balanced by the gravitations force of loop.

$$\begin{gathered} F_{net}=F_g-F \\ m\frac{dv}{dt}=mg-\frac{B{I}^{2}V}{R} \\ \frac{dv}{dt}=g-\frac{B{I}^2}{mR}v \end{gathered}$$

Substitute a for $\frac{B^2{I}^2}{mR}$into above equation.

$$\begin{gathered} \frac{dv}{dt}=g-\alpha v \\ \frac{dv}{g-\alpha v}=dt \end{gathered}$$

Integrate the above equation.

……. (3)

$\int \frac{dv}{g-\alpha v}=\int dt$

Let assume,

localid="1657621271021" $$\begin{gathered} g-\alpha v=u \\ 0-\alpha dv=du \\ dv=-\frac{du}{\alpha } \end{gathered}$$

Substitute for and for into equation (3).

$$\begin{gathered} \int -\frac{du}{u\alpha }=\int dt \\ \int -\frac{du}{u\alpha }=-a\int dt \\ In\left(u\right)=-\alpha t+InA \\ In\left(\frac{u}{A}\right)=-\alpha t \end{gathered}$$

Solve further as,

$$\begin{gathered} \frac{u}{A}=e^{-at} \\ u=A{e}^{at} \end{gathered}$$

Substitute g-av for u into above equation.

$$\begin{gathered} g-av=A{e}^{-at} \\ Att=0,v=0 \\ g-\left(0\right)a=A{e}^{-a\left(0\right)} \\ g=A \end{gathered}$$ …… (4)

Substitute the for into equation (4).

$$\begin{gathered} g-av=g{e}^{-at} \\ av=g-g{e}^{-at} \\ v=\frac{g}{\alpha }(1-e^{at}) \end{gathered}$$

Substitute $\alpha$ for$\frac{B^2{I}^2}{mR}$into above equation.

$$\begin{gathered} V=\frac{g}{{\displaystyle \frac{B^2{I}^2}{mR}}}\left(1-eat\right) \\ v=\frac{gmR}{B^2{I}^2}\left(1-eat\right) \end{gathered}$$ …… (5)

When the loop is move with the terminal velocity then the magnetic force is balanced by the gravitations force.

$$\begin{gathered} \frac{B^2{I}^2{v}_t}{R}=mg \\ {v}_t=\frac{mgR}{B^2{I}^2} \end{gathered}$$

…… (6)

Substitute$v_t$for$\frac{mgR}{B^2{I}^2}$into equation (5).

$v=v\left({}_{t}1-e^{-at}\right)$ …… (7)

The velocity at 90% of terminal velocity,$v=0.90v_t$

Substitute $0.90v_t$for v into equation (7).

$$\begin{gathered} 0.90v_t=v_t\left(1-e^{-at}\right) \\ 0.90=\left(1-e^{-at}\right) \\ {e}^{-at}=1-0.90 \\ t=-\frac{1}{a}In\left(0.1\right) \end{gathered}$$

Substitute $\alpha$for$\frac{B^2{I}^2}{mR}$into above equation.

$$\begin{gathered} t=-\frac{1}{{\displaystyle \frac{B^2{I}^2}{mR}}}In\left(0.1\right) \\ t=\frac{mR}{B^2{I}^2}n\left(10\right) \end{gathered}$$ …… (8)

The mass of the loop is given by,

$m=4nAI$

Here is the mass density of the aluminium, A is the cross-sectional area and i is the length of aluminium plate and$\sigma$is the conductivity of the aluminium.

The resistance of the loop is given by,

$R=\frac{4I}{A\sigma }$

Substitute$4nAI$for and $\frac{4I}{A\sigma }$for R into equation (8).

$$\begin{gathered} t=\frac{4nAI\times {\displaystyle \frac{4I}{A\sigma }}}{B^2{I}^2}In\left(10\right) \\ t=\frac{16np}{B^2}In\left(10\right) \end{gathered}$$

Substitute$2.7\times {10}^{3}kg/m^3$for n , $2.8\times {10}^{-10\Omega }$for P and 1 T for B into above equation.

$$\begin{gathered} t=\frac{16\times 2.7\times {10}^3\times 2.8\times {10}^{-8}}{1^2}In\left(10\right) \\ t=2.785\times {10}^{-3}s \\ t=2.785ms \end{gathered}$$

Recall equation (6)

$v_t=\frac{mgR}{B^2{I}^2}$

Substitute$4nAI$for and$\frac{4I}{A\sigma }$for R into above equation.

$$\begin{gathered} V_t=\frac{4nAI\times g\times {\displaystyle \frac{4I}{A\sigma }}}{B^2{I}^2} \\ {V}_t=\frac{16ngp}{B^2} \end{gathered}$$

Substitute for $2.7\times {10}^{3}kg/m^3$ for n , $2.8\times {10}^{-8\Omega }m$for$p,9.8m/s^2$ for g and 1 T I for Bnto above equation.

$$\begin{gathered} v_t=\frac{16\times 2.7\times {10}^3\times 9.8\times 2.8\times {10}^{-8}}{1^2} \\ {v}_t=0.01185m/s \end{gathered}$$

Hence the terminal velocity is$0.0185m/s$, velocity as function of time,$v_t\left(1-e^{-at}\right)$, and time taken to reach 90% of the terminal velocity is 2.8m/s .