# In which loop does the induced emf remain constant during the passage out of the magnetic field region?

A rectangular loop and a circular loop are moving out of a uniform magnetic field to a field-free region with a constant velocity V as shown in the figure. Explain in which loop do you expect the induced emf to be constant during the passage out of the field region. The magnetic field is normal to the loops.

Well if we take $B$ and $v$ as constants here then it is pretty easy, because all you need to do is calculate the change of the magnetic flux through the surface enclosed by either loops, and you know that the induced EMF ($\epsilon$) is given by: $$\epsilon=-N\frac{d\Phi}{dt}$$ where $\Phi = \mathbf{B}\cdot \mathbf{A}$, since we started off with $B$ constant then, $$\frac{d\Phi}{dt} \propto \frac{dA}{dt}$$ In the case of the rectangular loop, the amount by which the surface changes as the loop moves out, is constant, so: $$\frac{dA_{\rm rect}}{dt} = constant \leftrightarrow \epsilon = constant$$ Whereas the change of magnetic flux in the ciruclar loop isn't constant since the surface of the circle doesn't change by equal amounts as the loop moves out: $$\frac{dA_{\rm circ}}{dt}\ne constant$$
• @Swami No the magnetic flux through the moving circular loop isn't constant because the area change of the moving loop isn't constant. To elaborate a back of the envelope calculation shows: for the rectangle with sides L,x where x is parallel to $\vec{v}$, the area $A$ change per time is: $dA=L*dx=L*v*dt$ or in other words the derivative is constant: $\frac{dA_{rec}}{dt}=L*v.$ For the circle you have $\frac{d(\pi r^2)}{dt}=2*r*\pi*\frac{dr}{dt}$ which isn't constant. Hence the magnetic flux through the circular loop cannot be constant. – Ellie Aug 5 '14 at 8:15
• @Swami the radius of the whole loop is the same but the radius of the loop "inside" the $\mathbf{B}$ field is changing as the loop moves out. – Ellie Aug 5 '14 at 12:00
$$A = \frac{R^2}{2}(\theta - \sin\theta)$$