Briefly explain the change in flux due to rate of change of area?

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

In case of a conducting rod moving in 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 \(overrightarrow{dl}\)is moving with a velocity ‘V’ as shown in the figure.

In a time dt, \(overrightarrow{dl}\) sweeps out an area shown by ||elgram GFMN shown in the figure

\(,,,left| overrightarrow{da} right|=Vdtleft( dlsin theta right)\) \(,left| frac{overrightarrow{da}}{dt} right|=Vdlsin theta\) \(,,,frac{dvec{a}}{dt}=overrightarrow{dl}times overrightarrow{V}\) \(vec{B}.frac{dvec{a}}{dt}=vec{B}.left( overrightarrow{dl}times overrightarrow{V} right)\) \(,,,,,,,,,,,,,=left( overrightarrow{dl}times overrightarrow{V} right).vec{B}\) \(,,,,,,,,,,,,,=left( overrightarrow{V}times vec{B} right).overrightarrow{dl}\)

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

\(varepsilon =oint{left( vec{V}times vec{B} right)}.overrightarrow{dl}\)

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

Special care: If all parts of the loop are moving with uniform translational velocity ‘V” in an uniform magnetic filed ‘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.

\(varepsilon =oint{vec{V}times vec{B}}.overrightarrow{dl}\) \(,,,,=vec{V}times vec{B}.oint{overrightarrow{dl}=0}\)

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