Irodov 3.253: One Side Turns, Three Sides Weigh – Amit & Deepali Sinha Foundation

Problem Irodov 3.253

A copper wire (density $8.9\times 10^3$ kg/m $^3$ ) of cross-sectional area $S = 2.5$ mm $^2$ is bent into three sides of a square, a U-shaped frame whose two free ends are pivoted at $O$ and $O'$ on a horizontal axis, so that the frame hangs down from the axis and can swing about it. The frame is in a uniform vertical magnetic field. A current $I = 16$ A, led in and out along the axis, is passed through the wire, and the frame swings out from the vertical by the angle $\theta = 20^\circ$ . Find the magnetic induction.

The U-shaped copper frame pivoted on the horizontal axis OO', hanging vertically (solid) and swung out by θ when the current I flows (dashed), in the vertical field B. — The U-shaped copper frame pivoted on the horizontal axis $OO'$ , hanging vertically (solid) and swung out by $\theta$ when the current $I$ flows (dashed), in the vertical field $\vect B$ .

Given

$S = 2.5$ mm $^2$
$I = 16$ A
$\theta = 20^\circ$
copper, $\rho = 8.9\times 10^3$ kg/m $^3$ (table value)
$g = 9.8$ m/s $^2$

Find

the magnetic induction

B

Solution

Step 1Which forces turn the frame

Let each side of the square have length $a$ , so each has the mass

m = \rho S a \tag{1}

In a vertical field the Ampere force $I\,\vect a\times\vect B$ on a side is horizontal. On the two swinging sides it is parallel to the axis (they carry opposite currents, so these forces also cancel), and a force parallel to the axis has no torque about it. Only the bottom side, which runs parallel to the axis, feels a force that can turn the frame: $\vect a$ and $\vect B$ are perpendicular, so the force is horizontal, perpendicular to the axis, of magnitude

F = I a B \tag{2}

Gravity acts on all three sides.

Step 2The torques about the axis

The frame seen along the axis from the O end, swung out by θ: the weights 2m g of the two swinging sides and m g of the bottom side, and the horizontal Ampere force F on the bottom side, whose current runs into the page. — The frame seen along the axis from the $O$ end, swung out by $\theta$ : the weights $2m\vect g$ of the two swinging sides and $m\vect g$ of the bottom side, and the horizontal Ampere force $\vect F$ on the bottom side, whose current runs into the page.

Seen along the axis from $O$ , the current in the bottom side runs into the page. With $\vect i$ to the right, $\vect j$ up and $\vect k$ out of the page, the force on it is along

(-\vect k)\times\vect j = \vect i

horizontal and away from the vertical, which is the side the frame swings to. At the deflection $\theta$ the bottom side hangs a vertical distance $a\cos\theta$ below the axis, so $\vect F$ has the lever arm $a\cos\theta$ . The weights have the horizontal lever arms $(a/2)\sin\theta$ for each swinging side and $a\sin\theta$ for the bottom one. In equilibrium the torques balance, with (2):

I a B\cdot a\cos\theta = 2\cdot mg\,\frac{a}{2}\sin\theta + mg\,a\sin\theta = 2mga\sin\theta

I a B\cos\theta = 2mg\sin\theta

(3)

Step 3The field

Substituting the mass (1) into (3), the side length cancels:

I a B\cos\theta = 2\rho S a\,g\sin\theta

and, numerically,

B = \frac{2\,(8.9\times 10^3\text{ kg/m}^3)(9.8\text{ m/s}^2)(2.5\times 10^{-6}\text{ m}^2)\tan 20^\circ}{16\text{ A}}

B = \frac{2\rho g S}{I}\,\tan\theta \approx 9.9\text{ mT}

(4)

Check

Units: $\text{kg/m}^3\cdot\text{m/s}^2\cdot\text{m}^2$ over A is $\text{N}/(\text{A}\cdot\text{m})$ , a tesla. Energy route: the constant horizontal force $F$ on the bottom side has the potential energy $-Fa\sin\theta$ , and raising the three sides costs $mga(1 - \cos\theta)$ for the bottom and $2mg(a/2)(1 - \cos\theta)$ for the other two, $2mga(1 - \cos\theta)$ in all; setting the derivative of the sum with respect to $\theta$ to zero gives $Fa\cos\theta = 2mga\sin\theta$ , which is (3) with (2). Limit: the deflection can approach $90^\circ$ only as $B \to \infty$ , because the lever arm $a\cos\theta$ of the horizontal force shrinks to zero there while that of gravity grows to its largest.

Remark

The size of the frame drops out: both the magnetic force and the weight grow with the side $a$ , and both lever arms do too, so a frame of any size made of the same wire swings to the same angle. The balance therefore measures $B$ through the weight per unit length of the wire, $\rho S g$ , and the current: a cousin of the current balances that once realised the ampere by weighing magnetic forces.