Question

In Figure, a long rectangular conducting loop, of width L, resistance R, and mass m, is hung in a horizontal, uniform magnetic field$\overrightarrow{\mathbf{B}}$that is directed into the page and that exists only above line a. The loop is then dropped; during its fall, it accelerates until it reaches a certain terminal speed${\mathit{v}}_{\mathbf{t}}$. Ignoring air drag, find an expression for${\mathit{v}}_{\mathbf{t}}$.

Short answer

Terminal velocity of the loop is,$v_t=\frac{mgR}{B^2{L}^2}$

Step-by-step solution

Step 1: Given

i) Width of conducting loop, L

ii) Resistance of the loop , R

iii) Mass of the loop, m

iv) Uniform magnetic field going into the plane of paper,$\overrightarrow{B}$

Determining the concept

Use Faradays law of electromagnetic induction with Lenz law. The loop is moving in a uniform magnetic field so it experiences a force due to the applied magnetic field. This force must be balanced by the weight of the loop to achieve terminal velocity.

Faraday'slaw of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Lenz's law states that the current induced in a circuit due to a change in a magnetic field is directed to oppose the change in flux and to exert a mechanical force that opposes the motion

Formulae are as follow:

$$\begin{gathered} {\mathrm{\varphi }}_{\mathrm{B}}=\oint \overrightarrow{\mathrm{B}}.\mathrm{d}\overrightarrow{\mathrm{s}} \\ {\stackrel{,}{\mathrm{o}}}_{\mathrm{ind}}=\frac{{\mathrm{d\varphi }}_{\mathrm{B}}}{\mathrm{dt}} \\ \overrightarrow{\mathrm{F}}=\mathrm{id}\overrightarrow{\mathrm{I}}\times \overrightarrow{\mathrm{B}} \end{gathered}$$

Where,$\mathrm{\Phi }$is magnetic flux, B is magnetic field, i is current, 𝜀 is emf, l is length, F is force.

Determining the expression of for Vt

When the loop attains terminal velocity, its acceleration is zero. Therefore, forces acting on the loop are balanced. Therefore,

$\overrightarrow{F}=id\overrightarrow{I}\times \overrightarrow{B}=iL\left(-\hat{i}\right)\times B\left(-\hat{k}\right)=iLB\hat{j}=mg\hat{j}$

Assume y-axis to be parallel to the sides of the loop and x-axis to be parallel to the width of the loop.

$$\begin{gathered} iLB=mg \\ \therefore i=\frac{mg}{BL} \\ \stackrel{,}{{\mathrm{o}}_{\mathrm{ind}}=-\frac{{\mathrm{d\varphi }}_{\mathrm{B}}}{\mathrm{dt}}=-\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B}\left\{-\mathrm{dy}\right\}\mathrm{L}\right)=\mathrm{BL}.\frac{\mathrm{dy}}{\mathrm{dt}}}=BL{v}_t \end{gathered}$$

Here, dy is decreasing, so it is negative.

$$\begin{gathered} i=\frac{{{\displaystyle \stackrel{,}{o}}}_{ind}}{R}=\frac{BL{v}_t}{R}=\frac{mg}{BL} \\ \mathrm{Hence}, \\ \frac{{\mathrm{BLv}}_{\mathrm{t}}}{\mathrm{R}}=\frac{\mathrm{mg}}{\mathrm{BL}} \\ {\mathrm{v}}_{\mathrm{t}}=\frac{\mathrm{mgR}}{{\mathrm{B}}^2{\mathrm{L}}^2} \end{gathered}$$

Hence, terminal velocity of the loop is,${\mathrm{v}}_{\mathrm{t}}=\frac{\mathrm{mgR}}{{\mathrm{B}}^2{\mathrm{L}}^2}$

Therefore, Faraday’s law of electromagnetic induction and Lenz law is used to find out the emf induced in the loop.