Irodov 3.307: Growing Field Meets Growing Area – Amit & Deepali Sinha Foundation

Problem Irodov 3.307

A U-shaped conductor (two parallel bars joined at one end) lies in a uniform magnetic field perpendicular to its plane; the induction changes at the constant rate $\dot{B} = 0.10$ T/s. A conducting connector of length $l = 20$ cm lies across the bars and starts moving along them with acceleration $w = 10$ cm/s $^2$ . At $t = 0$ both the area of the circuit and the induction are zero. Neglecting the inductance of the circuit, find the emf induced in it at $t = 2.0$ s after the motion begins.

The U-shaped conductor, the connector of length l starting from the closed end with acceleration w, and the uniform field B normal to the plane. The sense of B is not given; drawn out of the page. — The U-shaped conductor, the connector of length $l$ starting from the closed end with acceleration $\vect w$ , and the uniform field $\vect B$ normal to the plane. The sense of $\vect B$ is not given; drawn out of the page.

Given

$\dot B = 0.10$ T/s, constant
$w = 10$ cm/s $^2$
$l = 20$ cm
$B = 0$ and area $0$ at $t = 0$
$t = 2.0$ s

Find

\mathcal{E}_i

t

Solution

Step 1The field and the area as functions of time

The induction grows from zero at the constant rate $\dot B$ . The connector starts from rest at the closed end (zero area at $t = 0$ ) and accelerates uniformly, so its distance $x$ from the closed end and the loop area $S = lx$ are:

\begin{aligned} B &= \dot B\,t \\ S &= l\,\frac{w t^2}{2} \end{aligned} \tag{1}

The circuit at time t: the area lx (washed), with x = wt^2/2, and the induced current I, clockwise for the field drawn out of the page and growing. — The circuit at time $t$ : the area $lx$ (washed), with $x = wt^2/2$ , and the induced current $I$ , clockwise for the field drawn out of the page and growing.

Step 2The flux and its rate of change

The field is uniform and normal to the loop, so the flux is the product of (1):

\Phi = BS = \frac{\dot B\,l\,w}{2}\,t^3

Faraday’s law then gives the emf’s magnitude from the time derivative:

\begin{aligned} |\mathcal{E}_i| &= \frac{d\Phi}{dt} = \frac{\dot B\,l\,w}{2}\cdot\frac{d}{dt}\,t^3 \\ &= \frac{\dot B\,l\,w}{2}\cdot 3t^2 \end{aligned}

Numerically, at $t = 2.0$ s,

|\mathcal{E}_i| = \frac{3}{2}\,(0.10\text{ T/s})(0.20\text{ m})(0.10\text{ m/s}^2)(2.0\text{ s})^2

|\mathcal{E}_i| = \frac{3}{2}\,w\,l\,\dot B\,t^2 = 12\text{ mV}

(2)

The field and the area both grow, so both add flux of the same sign; by Lenz’s rule the current circulates so as to oppose that growth, clockwise in the figure, where $\vect B$ is drawn out of the page.

Check

The emf splits into its two sources. The motional part, the connector moving at $v = wt$ through the field $B = \dot B t$ , is $Blv = \dot B l w t^2$ ; the transformer part, the changing field through the area (1), is $S\dot B = \dot B l w t^2/2$ . Their sum is $(3/2)\dot B l w t^2$ , matching (2): $8$ mV from the motion and $4$ mV from the changing field. Units: $\text{T/s}\cdot\text{m}\cdot\text{m/s}^2\cdot\text{s}^2 = \text{T}\,\text{m}^2/\text{s} = \text{V}$ .

Remark

Faraday’s flux rule covers both mechanisms at once: the Lorentz force on charges in the moving connector and the eddy electric field of the changing $B$ are physically different, yet $-d\Phi/dt$ counts them together. Here the motion contributes twice as much as the changing field at every instant, because $Blv$ uses the present velocity $wt$ , while the distance covered from rest, $x = wt^2/2$ , is only half of $wt$ times $t$ .