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

Equation (5.3.7) splits the emf into two parts with different origins, and each can occur without the other.

Motional emf

The circuit moves and $B$ is static. Figure 5.3.2 shows the basic example: a conducting bar of length $l$ slides with velocity $v = v a_{y}$ along two rails closed by a resistor $R$ , in a uniform field $B = - B a_{z}$ into the page. Only the bar moves, so only the bar contributes to the motional integral. On the bar

v \times B = v a_{y} \times (- B a_{z}) = - v B a_{x}, (5.3.8)

a force per unit charge directed along the bar (up the page), and integrating along the bar's length $l$ ,

e = \oint_{C} (v \times B) \cdot d l = B l v . (5.3.9)

Check with the flux rule: the loop encloses the area $l y (t)$ with $y = y_{0} + v t$ , so $Φ = B l y (t)$ and $∣ d Φ/ d t ∣ = B l v$ , the same answer. Since the flux into the page is growing, Lenz's law says the current circulates so as to make flux out of the page, that is counter-clockwise, up the bar and back through $R$ . Home exercise 5.3.2 completes the picture: the field pushes back on the current in the bar with the force $F = i l \times B$ , directed against $v$ , and the mechanical power needed to keep the bar moving is exactly the electrical power $e^{2} / R$ dissipated in the resistor. Drive the bar and the device is a generator, feed the rails from a source and the same bar is a linear motor (Exercise 5.3.4).

Motional emf. A bar slides to the right on two rails closed by a resistor, in a uniform field into the page. The magnetic force q\vec v\times\vec B on the charges in the bar drives a current i=Blv/R up the bar and round the loop. The field then acts on this current with the force \vec F=i\vec l\times\vec B, which opposes the motion. — **Figure 5.3.2.** Motional emf. A bar slides to the right on two rails closed by a resistor, in a uniform field into the page. The magnetic force $q v \times B$ on the charges in the bar drives a current $i = B l v / R$ up the bar and round the loop. The field then acts on this current with the force $F = i l \times B$ , which opposes the motion.

Change $B$ , $l$ , $v$ and $R$ , and reverse $v$ : the braking force always opposes the motion and $P_{mech} = P_{elec}$ .

InteractiveLoading simulation…

Transformer emf

The circuit is fixed and $B$ varies in time. Then $e = - \int_{S} (\partial B / \partial t) \cdot d S$ , and for a coil of $N$ close-wound turns $e = - N d Φ/ d t$ . This is the emf of the transformer secondary, of the induction coil and of every pick-up loop, worked out for a circular loop in a sinusoidal field in Home exercise 5.3.1 and for a loop beside a wire in Exercise 5.3.2. Problem 5.3.1 treats a frame in which both parts are present at once and shows that their sum is exactly $- d Φ/ d t$ .

Pause while $∣ B ∣$ grows and while it shrinks, and check the sense of the induced current.

InteractiveLoading simulation…

Recipe · Computing an induced emf

Orient. Choose $d S$ for the surface spanning the circuit and let the right hand fix the sense of $d l$ and the reference sense of $e$ and $i$ .
Flux. Compute $Φ (t) = \int_{S} B \cdot d S$ with the limits of the integral following the circuit if it moves. Only the component of $B$ along $d S$ counts.
Differentiate. $e = - d Φ/ d t$ . If the circuit changes shape or $B$ is not obviously a flux (the Faraday disc of Problem 5.3.2), use $e = \oint_{C} (v \times B) \cdot d l - \int_{S} (\partial B / \partial t) \cdot d S$ instead.
Check the sign with Lenz's law, then $i = e / R$ for a closed resistive loop.

Common mistake · Treating the induced field as a potential field

Writing $e = V_{A} - V_{B}$ and trying to assign a potential to the induced $E$ . The induced field has $\nabla \times E = - \partial B / \partial t \neq = 0$ , so $\oint_{C} E \cdot d l \neq = 0$ : the work done around a closed loop is not zero, which is exactly what an emf is, and no single-valued potential exists. A voltmeter connected across two points of such a loop reads a value that depends on how its leads are routed (Exercise 5.3.1).

Quick check

A fixed loop of area $0.01 m^{2}$ is placed with its normal along a uniform field that grows at $2 T/s$ . What is the emf, and in which sense does the induced current flow?

$∣ e ∣ = S d B / d t = 0.01 \times 2 = 20 mV$ (transformer emf). The current flows so that its own field inside the loop is opposite to the growing $B$ : counter-clockwise when seen looking along $B$ , that is clockwise when $B$ points towards the viewer, as in Figure 5.3.1.

Quick check

A $1 m$ rod moves at $10 m/s$ perpendicular to itself and to a $0.1 T$ field. What emf appears between its ends?

$e = B l v = 0.1 \times 1 \times 10 = 1 V$ (motional emf). Nothing need be closed for the emf to exist: the charges in the rod are pushed to one end until their field balances $v \times B$ .

5.3.4 The two ways to change the flux

Motional emf

Transformer emf