Irodov 3.288: A Chord Climbing a Parabola – Amit & Deepali Sinha Foundation

Problem Irodov 3.288

A wire is bent into the parabola $y = ax^2$ in the $xy$ plane, and a uniform magnetic field of induction $B$ is directed perpendicular to that plane (into the page as drawn). At $t = 0$ a straight connector, kept parallel to the $x$ axis, starts from rest at the apex of the parabola and moves up along the $y$ axis with a constant acceleration $w$ , closing a loop with the wire. Find the emf induced in this loop as a function of the connector’s height $y$ .

The parabolic wire y = ax^2 with apex O, the straight connector across it, accelerating up the y axis at w from rest at the apex, and the field B into the page. — The parabolic wire $y = ax^2$ with apex $O$ , the straight connector across it, accelerating up the $y$ axis at $w$ from rest at the apex, and the field $\vect B$ into the page.

Given

$y = ax^2$
$B$ normal to the plane, into the page
from rest at the apex
acceleration $w$

Find

\mathcal{E}_i(y)

Solution

Step 1The connector’s speed and length at height

y

The connector starts from rest at the apex and moves with constant acceleration $w$ , so when it has risen to height $y$ its speed follows from uniform acceleration kinematics:

v^2 = 2wy, \qquad v = \sqrt{2wy} \tag{1}

The part of the connector that belongs to the loop runs between the two branches, which it meets where the parabola reaches height $y$ . Solving the parabola’s equation for $x$ gives those ends and the chord’s length:

\begin{aligned} x &= \pm\sqrt{\frac{y}{a}} \\ l(y) &= 2\sqrt{\frac{y}{a}} \end{aligned} \tag{2}

The loop closed by the connector, its area (washed) growing as the connector rises, and the induced current I_ind it drives, counter-clockwise in the figure. — The loop closed by the connector, its area (washed) growing as the connector rises, and the induced current $I_{\mathrm{ind}}$ it drives, counter-clockwise in the figure.

Step 2The flux through the loop

The field is uniform and normal to the loop, so the flux is $B$ times the area under the chord. Summing horizontal strips of width $l(y')$ and height $dy'$ from the apex up to $y$ , with (2):

\begin{aligned} S(y) &= \int_0^{y} 2\sqrt{\frac{y'}{a}}\,dy' = \frac{4}{3}\,\frac{y^{3/2}}{\sqrt{a}} \\ \Phi(y) &= BS(y) = \frac{4B}{3\sqrt{a}}\,y^{3/2} \end{aligned}

Step 3Faraday’s law

Only $y$ changes, at the rate $dy/dt = v$ . Differentiating the flux by the chain rule:

\left|\frac{d\Phi}{dt}\right| = \frac{d\Phi}{dy}\,\frac{dy}{dt} = \frac{2B}{\sqrt{a}}\,y^{1/2}\,v = B\,l(y)\,v

The last form is the motional emf of the chord, $Blv$ , as it must be. Substituting (1):

\mathcal{E}_i = 2B\sqrt{\frac{y}{a}}\,\sqrt{2wy} = 2By\sqrt{\frac{2w}{a}}

\mathcal{E}_i = By\sqrt{\frac{8w}{a}}

(3)

The flux into the page grows, so by Lenz’s rule the induced current circulates counter-clockwise in the figure, its own field out of the page inside the loop.

Check

Dimensions: $y = ax^2$ makes $a$ an inverse length, so $w/a$ is $(\text{m/s}^2)(\text{m}) = \text{m}^2/\text{s}^2$ and $\sqrt{8w/a}$ is a speed; then $B\,y\sqrt{8w/a}$ is $\text{T}\cdot\text{m}\cdot\text{m/s} = \text{V}$ , as (3) requires. At the apex, $y = 0$ , both the chord and the speed vanish and so does the emf.

Remark

The speed grows only as $\sqrt{y}$ and the chord only as $\sqrt{y}$ , yet their product grows exactly as $y$ : the parabola’s widening and the uniform acceleration are matched. In time the height is $y = wt^2/2$ , so the emf rises as the square of the time:

\mathcal{E}_i = B\,\frac{wt^2}{2}\sqrt{\frac{8w}{a}} = Bt^2\sqrt{\frac{2w^3}{a}}

A steeper parabola (larger $a$ ) gives a narrower chord at every height and a smaller emf.