Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 5.3.1

Interactive: Faraday's Law of Induction

Problem 5.3.1

Moving frame in a time-varying field

A wire frame shown in the figure is on the $x = 0$ plane at $t = 0$ . Then it starts to move with a velocity of $v = v_{0} a_{y}$ . The magnetic flux density is measured as

B = B_{0} (sin y cos 2 t a_{x} + sin z sin 2 t a_{y}) [T] .

Find the electromotive force induced on the frame.

Adapted from: B. M. Notaroš, Electromagnetics, Problem 6.29.

Try it yourself first.

Strategy

The frame stays in the plane $x = 0$ , so its surface element is $d S = d S a_{x}$ and only the $a_{x}$ component of $B$ produces flux. Compute $Φ (t)$ with the moving limits $v_{0} t - a /2 \leq y \leq v_{0} t + a /2$ , then $e = - d Φ/ d t$ . As a check, compute the motional and the transformer terms of Eq. (5.3.10) separately: they must add up to the same answer.

1. The flux at time

t

At time $t$ the frame occupies $v_{0} t - a /2 \leq y \leq v_{0} t + a /2$ , $∣ z ∣ \leq b /2$ in the plane $x = 0$ (Figure 5.3.3a). The term $sin z sin 2 t a_{y}$ is parallel to the frame and contributes nothing to the flux. With $d S = d y d z a_{x}$ ,

Φ (t) = \int_{- b /2}^{b /2} \int_{v_{0} t - a /2}^{v_{0} t + a /2} B_{0} sin y cos 2 t d y d z = B_{0} b cos 2 t [- cos y]_{v_{0} t - a /2}^{v_{0} t + a /2} = B_{0} b cos 2 t [cos (v_{0} t - \frac{a}{2}) - cos (v_{0} t + \frac{a}{2})] .

Using $cos (A - B) - cos (A + B) = 2 sin A sin B$ with $A = v_{0} t$ and $B = a /2$ ,

Φ (t) = 2 B_{0} b sin \frac{a}{2} sin (v_{0} t) cos 2 t [Wb] . (5.3.11)

2. The emf

Differentiating the product $sin (v_{0} t) cos 2 t$ ,

e (t) = - \frac{d Φ}{d t} = - 2 B_{0} b sin \frac{a}{2} [v_{0} cos (v_{0} t) cos 2 t - 2 sin (v_{0} t) sin 2 t],

that is

Answer

e (t) = 2 B_{0} b sin \frac{a}{2} [2 sin (v_{0} t) sin 2 t - v_{0} cos (v_{0} t) cos 2 t] [V]

Check: term by term

The two terms in the bracket are exactly the two terms of Eq. (5.3.10). Transformer part: $(\partial B / \partial t) \cdot a_{x} = - 2 B_{0} sin y sin 2 t$ , so

- \int_{S} \frac{\partial B}{\partial t} \cdot d S = 2 B_{0} b sin 2 t \int_{v_{0} t - a /2}^{v_{0} t + a /2} sin y d y = 4 B_{0} b sin \frac{a}{2} sin (v_{0} t) sin 2 t .

Motional part: $v \times B = v_{0} a_{y} \times B_{0} sin y cos 2 t a_{x} = - v_{0} B_{0} sin y cos 2 t a_{z}$ has only a $z$ component, so only the two sides parallel to $z$ , at $y = v_{0} t \mp a /2$ , contribute. Going round the frame in the sense fixed by $d S = a_{x}$ (counter-clockwise seen from $+ x$ , that is $d l = d z a_{z}$ on the side $y = v_{0} t + a /2$ and $d l = - d z a_{z}$ on $y = v_{0} t - a /2$ ),

\oint_{C} (v \times B) \cdot d l = - v_{0} B_{0} b cos 2 t [sin (v_{0} t + \frac{a}{2}) - sin (v_{0} t - \frac{a}{2})] = - 2 v_{0} B_{0} b sin \frac{a}{2} cos (v_{0} t) cos 2 t .

The two parts add up exactly to the bracket above. Units: $B_{0} b sin (a /2)$ is $T \cdot m$ , and multiplied by the rate $2$ (in $s^{- 1}$ ) or by $v_{0}$ (in $m/s$ , with $sin y$ taken per metre) it gives $T \cdot m^{2} /s = Wb/s = V$ .

(a) The frame at time t. — (a) The frame at time $t$ .

(b) \Phi(t) with its envelope. — (b) $Φ (t)$ with its envelope.

(c) e(t) and its two parts. Problem 5.3.1. (a) The frame has moved to v_0t-a/2\le y\le v_0t+a/2 and its surface element is dS\,\vec a_x, so only B_x counts. (b), (c) With a=2, b=1 and v_0=1 (SI units) the factor 2\sin\frac a2=1.683. The flux is the product of a slow \sin v_0t, set by the *motion* through the pattern \sin y, and a fast \cos2t, set by the *source*: a beat between two independent rates. The emf is the sum of the transformer and the motional parts, which peak at different instants. — **Figure 5.3.3.** (c) $e (t)$ and its two parts. Problem 5.3.1. (a) The frame has moved to $v_{0} t - a /2 \leq y \leq v_{0} t + a /2$ and its surface element is $d S a_{x}$ , so only $B_{x}$ counts. (b), (c) With $a = 2$ , $b = 1$ and $v_{0} = 1$ (SI units) the factor $2 sin \frac{a}{2} = 1.683$ . The flux is the product of a slow $sin v_{0} t$ , set by the *motion* through the pattern $sin y$ , and a fast $cos 2 t$ , set by the *source*: a beat between two independent rates. The emf is the sum of the transformer and the motional parts, which peak at different instants.

Check: reading the graphs

Figures 5.3.3b and c show $Φ$ and $e$ for $a = 2$ , $b = 1$ , $v_{0} = 1$ . The flux vanishes whenever $v_{0} t = nπ$ (the frame sits symmetrically about a zero of $sin y$ ) or whenever $cos 2 t = 0$ . At those instants $e$ is not zero: the emf follows the slope of $Φ$ , not its value. Neither of the two contributions alone gives the envelope of the total.

Scrub through time and compare the transformer and motional parts with their sum.

InteractiveLoading simulation…

Problem 5.3.1: Moving frame in a time-varying field