Explain sign of emf using lenz's law

Here, the cause of the EMF is the motion of the rod to the right. The induced EMF establishes a current I in anticlockwise direction, so that the FB acting on the rod is to the left. Thus, opposing the motion of the rod in accordance with lenz’s law.

Direction of EMF alternate method:

As the rod is moving to the right

\(\begin{array}{l}-{{\phi }_{B}}\uparrow therefore \frac{d{{\phi }_{B}}}{dt}=-ve\end{array} \)

\(\begin{array}{l}\varepsilon =-\frac{d{{\phi }_{B}}}{dt}\end{array} \)

\(\begin{array}{l}\varepsilon =-\left( \frac{dv{{e}_{B}}}{dt} \right)\end{array} \)


Current is anticlockwise.

\(\begin{array}{l}\varepsilon =Blv\end{array} \)

In this case 3 vectors Blv one mutually ⊥ar the EMF induced is maximum.

Verification of conservation of energy:

If the resistance of the expanding loop is assumed as R, then the current flowing in the loop is
\(\begin{array}{l}i=\frac{\varepsilon }{R}=\frac{BLV}{R}\end{array} \)
The rate of W.D. by the ext. agency in moving the conducting rod with a constant velocity ‘v’ against the magnetic force
\(\begin{array}{l}{{F}_{B}}=ilB\end{array} \)
is given by

\(\begin{array}{l}P=(ILB)V\end{array} \)
\(\begin{array}{l}=\frac{{{B}^{2}}{{L}^{2}}{{V}^{2}}}{R}\end{array} \)

\(\begin{array}{l}=\frac{{{\varepsilon }^{2}}}{R}\end{array} \)
– power dissipated heat in overcoming the resistance ‘R’.

As the rod is moving on the parallel rails, smoothly ∴ one chemical energy is converted into electrical energy & which is dissipiated as heat is overcome the resistance of the loop.

Rate of change of magnetic flux due to rate of change of area (general case)

In case of a conducting rod moving on a π shaped conductor, only one side of the loop is moving. But now, we are considering the general case where all sides of the loop are moving in different direction with different velocities.

A part of the loop

\(\begin{array}{l}\overrightarrow{dl}\end{array} \)
is moving with a velocity ‘v’ as shown in the figure.


In a time dt,

\(\begin{array}{l}\overrightarrow{dl}\end{array} \)
sweeps out an area shown by parallelogram GFMN shown in the figure.

\(\begin{array}{l}\left| \overrightarrow{da} \right|=Vdt\left( dl\sin \theta \right)\end{array} \)

\(\begin{array}{l}\left| \frac{\overrightarrow{da}}{dt} \right|=Vdl\sin \theta\end{array} \)

\(\begin{array}{l}\frac{\overrightarrow{da}}{dt}=\overrightarrow{dl}\times \overrightarrow{V}\end{array} \)

\(\begin{array}{l}\overrightarrow{B}.\frac{\overrightarrow{da}}{dt}=\overrightarrow{B}.\left( \overrightarrow{dl}\times \overrightarrow{V} \right)\end{array} \)

\(\begin{array}{l}=\left( \overrightarrow{dl}\times \overrightarrow{V} \right).,\overrightarrow{B}\end{array} \)

\(\begin{array}{l}=\left( \overrightarrow{V}\times \overrightarrow{B} \right).,\overrightarrow{dl}\end{array} \)

If all sides of the loop are moving in different direction, then the induced EMF

\(\begin{array}{l}\varepsilon=\oint{\left( \overrightarrow{V}\times \overrightarrow{B} \right)}\overrightarrow{dl}\end{array} \)

This is the general formula for induced EMF due to time varying area of the loop.

Special case:

If all parts of the loop are moving with uniform translational velocity of in an uniform magnetic field ‘B’, then the loop observes no change in magnetic flux. Passing through it & hence the induced emf is equal to zero.

It is similar to the rigid loop moving in a uniform magnetic field

∴same flux passes through it.

\(\begin{array}{l}\varepsilon =\oint{\overrightarrow{V}\times \overrightarrow{B}}.,\overrightarrow{dl}\end{array} \)
\(\begin{array}{l}=\overrightarrow{V}\times \overrightarrow{B}.\oint{\overrightarrow{dl}}=0\end{array} \)

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