# Rate of change of area of a circular loop exiting a uniform magnetic field

The question is to find the variation in induced electromotive force w.r.t to time or put simply $$\varepsilon(t)$$ for a circular loop being removed from the region of the magnetic field at a constant velocity $$v$$.
Obviously $$\varepsilon= \frac{d\phi}{dt}$$ Now $$\phi = BA\cos \theta=B(\pi r^2)$$ Now, $$\varepsilon= \frac{d\phi}{dt} = 2\pi rB \frac{dr}{dt}$$ But there's no clear way to determine $$\frac{dr}{dt}$$. So is there any other way to determine the rate of change of area (how much of the circle is leaving the region per $$dt$$ time). Any help would be very good right now.

• Maybe try using $v=\frac{dr}{dt}$, and assume that the loop was initially centered at $r=0$? – amateurAstro Jul 17 at 19:41
• but according to that, expression the area changing is constant? – Prakhar Nagpal Jul 17 at 19:43
• The radius of the loop doesn't change so $dr/dt = 0$. You need to express the area of the loop through which a (constant?) magnetic field is passing as a function of time, and differentiate that. Note also that strictly speaking a magnetic field abruptly decreasing to zero is unphysical. – Puk Jul 17 at 19:45
• @PrakharNagpal, If the field is uniform everywhere, then how can you remove the loop from the magnetic field? Moving a loop through a uniform field will not induce any emf. It seems the problem as you describe it is not well posed. – amateurAstro Jul 17 at 19:49
• @amateurAstro the field is uniform in a certain region and the loop is being pulled out from that region at a constant velocity then obviously since the area is decreasing non uniformly resulting in a change in flux! If you feel like the question can be edited please feel free to do so with reference to this comment. – Prakhar Nagpal Jul 17 at 20:05

As I mentioned in a comment, a magnetic field abruptly decreasing to zero is unphysical, unless you have currents or time-varying E-fields around it to confine it. However I will ignore this concern for what follows.

If the magnetic field is confined to a large rectangular region in space, when the loop is leaving this region the magnetic field is passing through only a portion of the loop. This portion has the shape of the intersection of a circle and a half-plane. Let the edge of this half-plane be located a distance $$x$$ from the center of the circle. The secant line defined by the boundary of the region with magnetic field has length $$2\sqrt{r^2-x^2}$$. The rate of change of the area enclosing the magnetic field with $$x$$ is this secant length. If the loop is halfway outside the magnetic field at $$t=0$$, we can write $$x=-vt$$ where $$v$$ is the speed. Putting everything together,

$$\mathcal{E} = -\frac{d\Phi}{dt} = -B \cos\theta \frac{dA}{dt} = -B \cos\theta \frac{dx}{dt} \frac{dA}{dx} = 2B v \sqrt{r^2 - (vt)^2} \cos\theta$$

for $$|t| < r/v$$, and $$0$$ for all other times.

• So if I had to plot a graph of $\varepsilon$ vs $t$ it according to this it would be a semicircle? – Prakhar Nagpal Jul 17 at 20:43
• It would be a semi-ellipse, with semi-major axes determined by the various parameters involved. – Puk Jul 17 at 20:45
• Specifically the principal semi-axis along the $\mathcal{E}$ axis is $2Bvr \cos \theta$ and the semi-axis along $t$ is $r/v$. – Puk Jul 17 at 20:57
• Are you sure that the $\cos \theta$ term should be there? – Prakhar Nagpal Jul 18 at 4:48
• It should be there if $\vec{B}$ is not normal to the loop normal vector. In this case $\theta$ is the angle between these vectors. – Puk Jul 18 at 5:02

What you've derived is a formula for the emf generated in a single turn coil whose radius is changing at a rate $$\frac{dr}{dt}$$. The coil's axis is parallel to a constant uniform field of magnitude B.

But you're trying to find an emf in a coil of fixed radius? You can't, I'm afraid, find how the emf varies for "a circular loop being removed from the region of the magnetic field at a constant velocity 𝑣" unless you know how the field varies from point to point, and the path along which you move the loop. Even if you do know these things it may still be hard to do what you want.