The conducting rod shown in Figure has length L and is being pulled along horizontal, frictionless conducting rails at a constant velocity$\overrightarrow{\mathbf{v}}$. The rails are connected at one end with a metal strip. A uniform magnetic field$\overrightarrow{\mathit{B}}$, directed out of the page, fills the region in which the rod moves. Assume that$\mathit{L}\mathbf{=}\mathbf{10}\mathit{c}\mathit{m}\mathbf{,}\mathit{v}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\frac{\mathbf{m}}{\mathbf{s}}\mathbf{,}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{B}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{}\mathit{T}$. (a) What is the magnitude of the emf induced in the rod? (b) What is the direction (up or down the page) of the emf induced in the rod? (c) What is the size of the current in the conducting loop? (d) What is the direction of the current in the conducting loop? Assume that the resistance of the rod is 0.40 and that the resistance of the rails and metal strip is negligibly small. (e) At what rate is thermal energy being generated in the rod? (f) What external force on the rod is needed to maintain$\overrightarrow{\mathbf{v}}$? (g) At what rate does this force do work on the rod?

1. Length of conducting rod ,$L=10.0cm=10\times {10}^{-2}m$
2. Uniform magnetic field coming out of page$\overrightarrow{B}=B_0\hat{k}$,$B_0=1.2T$
3. Velocity of conducting rod$\overrightarrow{v}=-v_0\hat{i}$,$v_0=5.0\mathrm{m}/\mathrm{s}$
4. Resistance of the rod is$R=0.40\Omega$

Use Faraday’s law of electromagnetic induction with Lenz law. The rod is moving in a uniform magnetic field, so it experiences a force due to the applied magnetic field. Also, mechanical power is the dot product of force and 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 follows:

${\varphi }_B=\oint \overrightarrow{B}·d\overrightarrow{s}$

$$\begin{gathered} {\stackrel{,}{O}}_{ind}=-\frac{d{\varphi }_B}{dt} \\ \overrightarrow{F}=i\overrightarrow{dl}\times \overrightarrow{B} \\ P=i^{2}R \\ P=\overrightarrow{F}·\overrightarrow{v} \end{gathered}$$

Where,$\Phi$is magnetic flux, B is magnetic field, i is current, R is resistance,𝜀 is emf, l islength, F is force, v is velocity, P is power

$d{\varphi }_B=\overrightarrow{B}·d\overrightarrow{A}=B_0\hat{k}\left(-dA\right)\hat{k}=-B_{0}dA$

Where,$dA=dx·L$

$$\begin{gathered} {\stackrel{,}{O}}_{ind}=-\frac{d{\varphi }_B}{dt} \\ {\stackrel{,}{O}}_{ind}=-\frac{d}{dt}\left(-B_{0}dA\right) \\ {\stackrel{,}{O}}_{ind}=-\frac{d}{dt}{B}_{0}dx·L \\ {\stackrel{,}{O}}_{ind}=B_0·L·\frac{dx}{dt} \\ {\stackrel{,}{O}}_{ind}=B_{0}L{v}_0 \\ {\stackrel{,}{O}}_{ind}=B_{0}Lv=1.2\times 10\times {10}^{-2}\times 5.0 \\ {\stackrel{,}{O}}_{ind}=0.60V \end{gathered}$$

Hence, the magnitude of induced emf is0.60 V.

By Lenz law, as the applied magnetic field is along z-axis, to oppose that, the induced magnetic field will be along negative z – axis. Therefore, the emf must be directed up the page.

Hence, the direction of induced emf is upside.

By using Ohms law, findthe current throughthe rod.

$$\begin{gathered} i_{ind}=\frac{{{\displaystyle \stackrel{,}{O}}}_{ind}}{R} \\ {i}_{ind}=\frac{0.60}{0.40}V \\ {i}_{ind}=1.5A \end{gathered}$$

Hence, the value of current is 1.5 A.

Direction ofthe current is determined by the direction of emf, so it is clockwise.

Hence, the direction of current is clockwise.

The rate at which heat is generated in the rod is electrical power, so ,

$$\begin{gathered} P=i_{ind}^{2}R \\ P=1.5\times 1.5\times 0.40 \\ P=0.90W \end{gathered}$$

Hence, the rate of thermal energy is 0.90 W.

The force acting onthe current loop kept in magnetic field is,$\overrightarrow{F}=i\overrightarrow{dl}\times \overrightarrow{B}$

$\overrightarrow{F}=i\overrightarrow{dl}\times \overrightarrow{B}=i_{ind}L\hat{j}\times B_0\hat{k}=i_{ind}L{B}_0\hat{i}$

So thevalue of external force to maintain

${\overrightarrow{F}}_{ext}=i_{ind}L{B}_0\hat{i}=-1.5\times 10\times {10}^{-2}\times 1.2\hat{i}=-0.18\hat{i}N$

Hence, magnitude of applied force is, role="math" localid="1661918931404" $F_{ext}=-0.18N$.

Rate of external work:

Mechanical power

$$\begin{gathered} P=\overrightarrow{F_{ext}}·\overrightarrow{v} \\ P=\left(-0.18\hat{i}\right)·\left(-v_0\hat{i}\right) \\ P=0.18\times 5.0 \\ P=0.90W \end{gathered}$$

Hence, rate of external work done is 0.90 W.

Therefore, the rate of external work is nothing but the power delivered to the specific process. In this case, to move the rod, power is spent by the external force.