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# Induced e.m.f is given by it will be maximum when,

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## Faraday's Law of Electromagnetic InductionInduced emf in a circuit is proportional to the rate of change of magnetic flux linked with the circuit.The direction of induced emf is such that it tries to oppose the cause of its generation, i.e., the variation of magnetic flux producing it.Induced EMFIf $\varphi$ be the magnetic flux linked with a circuit at any time $t$ then the laws of electromagnetic induction can be expressed mathematically as$\epsilon =-\frac{d\varphi }{dt}$, $\left(1\right)$Where $\epsilon$ is the induced emf. Here it is important to point out that $\epsilon$ does not depend on how the flux $\varphi$ is changed.If the electric field in space is denoted by $\stackrel{\to }{E}$ then, by definition, the emf around a closed path $C$ is$\epsilon ={\oint }_{C}\stackrel{\to }{E}.\stackrel{\to }{dl}.$If $S$ is an open surface bounded by the curve $C$ placed in a magnetic field $\stackrel{\to }{B}$ then the magnetic flux through the surface is$\varphi ={\int }_{S}\stackrel{\to }{B}.d\stackrel{\to }{S.}$So equation $\left(1\right)$ can be written as${\oint }_{C}\stackrel{\to }{E}.d\stackrel{\to }{l}=-\frac{d}{dt}{\int }_{S}\stackrel{\to }{B}.d\stackrel{\to }{S}.$ $\left(2\right)$This is the integral form of Faraday's law.Therefore, from equation $\left(2\right)$ we can see that for maximum emf, the angle between $\stackrel{\to }{B}$ and $\stackrel{\to }{S}$ must be $0°$.Hence, for maximum induced emf the open surface must be parallel to the magnetic field.  Suggest Corrections  0      Similar questions  Explore more