Irodov 3.254: A Couple Weighed on a Balance – Amit & Deepali Sinha Foundation

Problem Irodov 3.254

A balance beam pivots about a horizontal axis at $O$ . A pan hangs from the end $A$ of one arm, of length $OA = l = 30$ cm. The other end of the beam carries a small coil $C$ of $N = 200$ turns and cross-sectional area $S = 1.0$ cm $^2$ , its axis along the beam, placed in the gap between the north (upper) and south (lower) poles of an electromagnet. With no current in the coil the balance is level. When a current $I = 22$ mA is passed through the coil, the balance is brought back to level by adding a counterweight of mass $\Delta m = 60$ mg to the pan. Find the magnetic induction at the coil.

The balance: the coil C on one end of the beam, its axis along the beam, in the gap between the poles N and S; the pan with the counterweight Δ m hangs at A, a distance l from the pivot O. — The balance: the coil $C$ on one end of the beam, its axis along the beam, in the gap between the poles N and S; the pan with the counterweight $\Delta m$ hangs at $A$ , a distance $l$ from the pivot $O$ .

Given

$N = 200$
$S = 1.0$ cm $^2$
$l = OA = 30$ cm
$I = 22$ mA
$\Delta m = 60$ mg
field uniform over the small coil (assumed)

Find

the induction

B

at the coil

Solution

Step 1What the field does to the coil

The gap field is vertical, from the N pole down to the S pole, and the coil is small, so the field is uniform over it (assumed). A uniform field exerts no net force on a closed current loop, only a couple. The moment of the coil is along its axis, which lies along the beam, so it is perpendicular to $\vect B$ :

p_m = N I S \tag{1}

and the couple on it has the magnitude

\tau = |\vect p_m \times \vect B| = p_m B = N I S B \tag{2}

The beam as a lever: the coil's moment p_m along the beam, the gap field B across it, the couple τ that turns the beam about O, and the counterweight's weight Δ m g at A, a lever arm l from O. — The beam as a lever: the coil’s moment $\vect p_m$ along the beam, the gap field $\vect B$ across it, the couple $\vect\tau$ that turns the beam about $O$ , and the counterweight’s weight $\Delta m\,\vect g$ at $A$ , a lever arm $l$ from $O$ .

The couple $\vect\tau = \vect p_m\times\vect B$ is horizontal and normal to the vertical plane in which the beam swings, so it turns the beam about the pivot axis. Here it must lift the pan’s side, which is why weight is added there: with $\vect B$ pointing down, the right-hand rule then requires $\vect p_m$ to point along the beam away from $O$ , as drawn.

Step 2Balancing the couple

The beam is level again when the moment of the counterweight’s weight about $O$ equals the couple (2):

\Delta m\,g\,l = N I S B

Numerically,

B = \frac{(60\times10^{-6}\text{ kg})(9.8\text{ m/s}^2)(0.30\text{ m})}{(200)(22\times10^{-3}\text{ A})(1.0\times10^{-4}\text{ m}^2)}

B = \frac{\Delta m\,g\,l}{N I S} = 0.40 \text{ T}

(3)

Check

Units: $\text{kg}\cdot(\text{m/s}^2)\cdot\text{m}$ is a $\text{N}\cdot\text{m}$ , and $\text{N}\cdot\text{m}/(\text{A}\cdot\text{m}^2) = \text{N}/(\text{A}\cdot\text{m}) = \text{T}$ . Independence of the coil’s arm: sliding the coil along the beam, still inside the uniform gap field, leaves the couple (2) and so the balance unchanged, as a couple must, which is why only the pan’s arm $l$ appears in the answer.

Remark

Reversing the current reverses $\vect p_m$ and the couple, so the counterweight would then go on the coil’s side. That is the instrument’s virtue: it weighs the couple $N I S B$ , in which only the coil’s geometry and the measured current appear, so it is an absolute way to measure $B$ , the principle of the classic current balance.