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

Home exercise 5.3.2

Interactive: Faraday's Law of Induction

Home exercise 5.3.2

Sliding bar on rails

A conducting bar of length $l$ slides at constant speed $v$ on two parallel rails closed by a resistor $R$ , in a uniform magnetic flux density $B$ perpendicular to the plane of the rails. The bar and the rails have negligible resistance.

Find the emf $e$ and the current $i$ in the circuit.
Find the magnetic force $F$ on the bar and its direction.
Find the mechanical power needed to keep the bar moving and compare it with the electrical power dissipated in $R$ .

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 7-2.

Try it yourself first.

Strategy

The generator of Section 5.3.4. Only the bar moves, so the motional emf is $\int_{bar} (v \times B) \cdot d l$ . The current then feels the force $i l \times B$ , and the power balance closes the argument.

Take the plane of the rails as the $x y$ -plane with the bar moving along $+ a_{y}$ (to the right in the figure), $x$ down the page and $B = - B a_{z}$ (into the page), as in Figure 5.3.2.

Only the bar moves, so only it contributes to the motional emf:

v \times B e = v a_{y} \times (- B a_{z}) = - v B a_{x}, = \oint_{C} (v \times B) \cdot d l = \int_{bar} (- v B a_{x}) \cdot d l = B l v,

taken along the bar in the $- a_{x}$ direction (up the page). Check with the flux rule: the loop area is $l y (t)$ , $Φ = B l y$ and $∣ d Φ/ d t ∣ = B l v$ . The current is

Answer

e = B l v, i = \frac{e}{R} = \frac{B l v}{R}

It flows up the bar (the sense of $v \times B$ ) and round the loop through $R$ . Lenz: the loop area, and hence the flux into the page, is growing, so the current circulates counter-clockwise to produce flux out of the page.

The force on the current-carrying bar is $F = i l \times B$ with $l = - l a_{x}$ (the direction of the current):

Answer

F = i (- l a_{x}) \times (- B a_{z}) = - i l B a_{y} = - \frac{B ^{2} l ^{2} v}{R} a_{y}

It is directed against $v$ : the field brakes the bar.

To keep $v$ constant an external force $- F$ must be applied, delivering the mechanical power

P_{mech} while P_{elec} = ∣ F ∣ v = \frac{B ^{2} l ^{2} v ^{2}}{R}, = i^{2} R = \frac{e ^{2}}{R} = \frac{B ^{2} l ^{2} v ^{2}}{R} .

Answer

P_{mech} = P_{elec} = \frac{B ^{2} l ^{2} v ^{2}}{R}

The magnetic field stores nothing here. It only converts mechanical work into heat in $R$ .

The defaults are the numbers of the Check: read off 1.2 V, 2.4 A, 0.36 N and 2.88 W.

InteractiveLoading simulation…

Check

Numbers: $B = 0.60 T$ , $l = 0.25 m$ , $v = 8 m/s$ , $R = 0.5 Ω$ give $e = 1.2 V$ , $i = 2.4 A$ , $F = 0.36 N$ and $2.88 W$ on both sides. Three laws meet in one picture: $F = q v \times B$ makes the emf, $J = σ E$ turns it into a current, and $F = i l \times B$ turns the current back into a force (all three from Chapter 4). Drive the bar and it is a generator. Feed $R$ from a source instead and the same bar is a linear motor (Exercise 5.3.4). Released without an applied force, the bar coasts to rest exponentially with $τ = m R / (B^{2} l^{2})$ .

Home exercise 5.3.2: Sliding bar on rails