# Find Induced Emf in this Coil

by ltkach
Tags: induced emf, magnetism
 Sci Advisor Thanks P: 2,440 The electromotive force is 0 here. So your calculation is correct. This can also be immediately seen from Faraday's Law in integral form. To derive it one must be a bit carefull. It's wrong in many textbooks. One always should start from the local (differential) form of the Lorentz invariant Maxwell equations. The Faraday Law in any inertial frame of reference reads $$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$ Now let $A$ be any area (time dependent or not) and $\partial A$ its boundary. Further let the relative orientation of the surface elements and the boundary curve be defined in the standard way by the right-hand rule. Then, according to Stokes's theorem you have $$\int_{\partial A} \mathrm{d} \vec{x} \cdot \vec{E}=-\int_A \mathrm{d} \vec{A} \partial_t \vec{B}. \qquad (*)$$ Now the point is to bring the partial time derivative out of the integral. This, however gives two contributions: Besides the partial time derivative of the time-dependent magnetic field, it also contains a piece from the change of the moving surface. The result of the analysis is $$\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d} \vec{A} \cdot \vec{B}=\int_A \mathrm{d} \vec{A} \cdot \partial_t \vec{B}-\int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{B}).$$ Here $\vec{v}=\vec{v}(t,\vec{r})$ is the velocity of the boundary-line element at the time and position $(t,\vec{r})$. The proof is given in the Wikipedia: http://en.wikipedia.org/wiki/Faraday...araday.27s_law You have to open the box to see the proof. Now we have the magnetic flux defined by the integral on the left-hand side, $$\Phi(t)=\int_{A} \mathrm{d} \vec{A} \cdot \vec{B},$$ and then Faraday's Law (*) tells us $$\int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B})=-\frac{\mathrm{d} \Phi}{\mathrm{d} t}.$$ In your case, because the magnetic field is homogeneous, the magnetic flux through the coil doesn't change with time and thus the EMF, i.e., the line integral on the left-hand side of the above equation, vanishes.