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A flat spiral coil is wound from thin insulated wire in one plane, like the heating coil of an…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A flat spiral coil is wound from thin insulated wire in one plane, like the heating coil of an induction cooker. It has $20$ turns, spread evenly from an inner radius of $2.0\ \text{cm}$ to an outer radius of $6.0\ \text{cm}$ , as in the figure. It carries a steady current of $3.0\ \text{A}$ . The two leads run close together, so they add no field. The turns are so close that you may treat the coil as $20$ circular turns, with the same number of turns in each centimetre of radius. Its centre is $O$ , and its axis passes through $O$ at right angles to its plane. Take $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$ .

A flat spiral coil of 20 turns seen from above, wound from an inner radius of 2.0 cm out to an outer radius of 6.0 cm about its centre O. The two leads run side by side from the two ends of the spiral to the right. The current is 3.0 A. A note says the axis passes through O at right angles to the page.

What is the magnetic field at $O$ ?
What is the field on the axis, $4.0\ \text{cm}$ from $O$ ?
What is the magnetic moment of the coil?
What is the field on the axis $60\ \text{cm}$ from $O$ , where the coil acts as a small magnetic dipole?
The coil is placed in a uniform field of $50\ \text{mT}$ that makes $30^\circ$ with the plane of the coil. What torque acts on it?
How much work is needed to turn the coil slowly from its stable position in that field to its unstable one?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Treat the coil as many thin rings, each with its own radius, and add their fields and moments with integrals. Far away the whole coil is a dipole.

Step 1. Turns per metre of radius. $n = \dfrac{N}{b - a} = \dfrac{20}{0.04\ \text{m}} = 500\ \text{m}^{-1}$ .

Step 2. Field at the centre (a).

\displaystyle B_O = \int_a^b \frac{\mu_0 I n}{2r}\,dr = \frac{\mu_0 N I}{2(b - a)}\ln\frac{b}{a} = 1.035\ \text{mT}

Step 3. Field on the axis (b). With $\displaystyle\int \frac{r^2\,dr}{(r^2 + z^2)^{3/2}} = \sinh^{-1}\frac{r}{z} - \frac{r}{\sqrt{r^2 + z^2}}$ ,

\displaystyle B(z) = \frac{\mu_0 I n}{2}\left[\sinh^{-1}\frac{r}{z} - \frac{r}{\sqrt{r^2 + z^2}}\right]_a^b = 0.310\ \text{mT}

at $z = 4.0\ \text{cm}$ .

Step 4. Moment (c). $m = \displaystyle\int_a^b I\pi r^2 n\,dr = \frac{\pi N I(a^2 + ab + b^2)}{3} = 0.327\ \text{A m}^2$ .

Step 5. Far away (d). On the axis of a dipole, $B = \dfrac{\mu_0 m}{2\pi z^3} = \dfrac{4\pi \times 10^{-7}\ \text{T m/A} \times 0.327\ \text{A m}^2}{2\pi(0.60\ \text{m})^3} = 0.303\ \mu\text{T}$ .

Step 6. Torque and work (e, f). $\vec m$ is at right angles to the plane, so it makes $60^\circ$ with the field: $\tau = mB\sin60^\circ = 0.0141\ \text{N m}$ . From $\vec m$ along $\vec B$ to $\vec m$ against it, the energy $-mB\cos\theta$ rises by $2mB = 0.0327\ \text{J}$ .

Answer: (a) $1.035\ \text{mT}$ . (b) $0.310\ \text{mT}$ . (c) $0.327\ \text{A m}^2$ . (d) $0.303\ \mu\text{T}$ . (e) $0.0141\ \text{N m}$ . (f) $0.0327\ \text{J}$ .

Watch out. Putting all $20$ turns at the mean radius of $4.0\ \text{cm}$ gives $0.942\ \text{mT}$ at the centre, too small: the inner turns, being closer, count for more. And the angle in the torque is the one between $\vec m$ and $\vec B$ , $60^\circ$ here, not the $30^\circ$ with the plane.

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