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Question-42278




Question Number 42278 by rish@bh last updated on 21/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18
excellent problem...  Ψ=B^→ .A^→   (dΨ/dt)=B^→ .(dA^→ /dt)+(dB^→ /dt).A^→       B=magnetic flux  A^→ =area vector  here(dA^→ /dt)=0since no relative motion between B^→   and area vector A^→     (dΨ/dt)=(dB/dt).A^→   −e=(dΨ/dt)=(dB^→ /dt).A^→   now what is the value of A^→   let me think...wait...
$${excellent}\:{problem}… \\ $$$$\Psi=\overset{\rightarrow} {{B}}.\overset{\rightarrow} {{A}} \\ $$$$\frac{{d}\Psi}{{dt}}=\overset{\rightarrow} {{B}}.\frac{{d}\overset{\rightarrow} {{A}}}{{dt}}+\frac{{d}\overset{\rightarrow} {{B}}}{{dt}}.\overset{\rightarrow} {{A}}\:\:\:\:\:\:{B}={magnetic}\:{flux} \\ $$$$\overset{\rightarrow} {{A}}={area}\:{vector} \\ $$$${here}\frac{{d}\overset{\rightarrow} {{A}}}{{dt}}=\mathrm{0}{since}\:{no}\:{relative}\:{motion}\:{between}\:\overset{\rightarrow} {{B}} \\ $$$${and}\:{area}\:{vector}\:\overset{\rightarrow} {{A}} \\ $$$$ \\ $$$$\frac{{d}\Psi}{{dt}}=\frac{{dB}}{{dt}}.\overset{\rightarrow} {{A}} \\ $$$$−{e}=\frac{{d}\Psi}{{dt}}=\frac{{d}\overset{\rightarrow} {{B}}}{{dt}}.\overset{\rightarrow} {{A}} \\ $$$${now}\:{what}\:{is}\:{the}\:{value}\:{of}\:\overset{\rightarrow} {{A}} \\ $$$${let}\:{me}\:{think}…{wait}… \\ $$

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