Chapter 4 · Moving Charges and Magnetism

4.4 Torque on a Current Loop and the Moving-Coil Galvanometer

A current loop in a magnetic field feels no net force but a torque $\boldsymbol\tau=\mathbf m\times\mathbf B$ that turns it like a compass needle. Balance that torque against a spring and you have the moving-coil galvanometer — and, with one extra resistor, an ammeter or a voltmeter.

Every electric motor, from the one that spins a ceiling fan to the ones that drive a metro train, works because a coil carrying a current in a magnetic field is turned. Before digital meters, the same effect was used to measure current: in an analogue multimeter a small coil, hung in the field of a permanent magnet, twists against a fine spring until the two torques balance, and a pointer fixed to the coil shows how far it has turned. Many school laboratories and car dashboards still use such moving-coil meters, and the physics behind them is the subject of this section.

In Section 4.1 we found the force on a straight wire, and saw that the forces on a closed loop in a uniform field add to zero. Here we show that those forces nevertheless form a couple, find the torque on a current loop, and meet the quantity that controls it — the loop's magnetic dipole moment. We then see that a small current loop behaves, at a distance, exactly like a magnetic dipole, and finish with the moving-coil galvanometer and how to convert it into an ammeter or a voltmeter.

Torque on a Current Loop in a Uniform Magnetic Field

Take a rectangular loop $PQRS$ with sides $PQ=RS=a$ and $QR=SP=b$, carrying a steady current $I$, in a uniform field $\mathbf B$. The loop can turn about an axis through its centre parallel to the sides of length $b$, and this axis is perpendicular to $\mathbf B$.

Plane of the loop parallel to $\mathbf B$. Suppose the sides of length $a$ lie along $\mathbf B$. They feel no force ($\sin0^\circ=0$). The sides of length $b$ are perpendicular to $\mathbf B$ and carry the current in opposite directions, so they feel equal and opposite forces of size $IbB$, perpendicular to the plane of the loop. These two forces act along different lines, a distance $a$ apart: they form a couple whose torque is $$\tau=IbB\cdot a=IAB,$$ where $A=ab$ is the area of the loop. The couple turns the loop so that its plane swings round to face the field.

Plane of the loop at an angle. Now let the normal to the loop make an angle $\theta$ with $\mathbf B$. The forces on the sides of length $a$ are equal, opposite and act along the same line (the rotation axis), so they cancel and give no torque. The forces on the sides of length $b$ are still $IbB$ each (those sides are still perpendicular to $\mathbf B$), but the perpendicular distance between their lines of action is now $a\sin\theta$. So $$\tau=IbB\cdot a\sin\theta=IAB\sin\theta,$$ and with $N$ turns, each adding the same torque, $\tau=NIAB\sin\theta$.

📘 Definition 4.5 — Magnetic dipole moment of a current loop
A plane loop of area $A$ with $N$ turns, each carrying a current $I$, has magnetic dipole moment (magnetic moment) $$\mathbf m=NIA\,\hat{\mathbf n},\qquad m=NIA,$$ where $\hat{\mathbf n}$ is the unit normal to the loop given by the right-hand thumb rule (fingers along the current, thumb along $\hat{\mathbf n}$) — the same direction as the field the loop produces at its centre. The SI unit of $m$ is the ampere square metre, $\text{A m}^2$ (equivalently $\text{J/T}$).
📐 Result 4.11 — Force and torque on a current loop in a uniform field

A plane current loop of magnetic moment $\mathbf m$ in a uniform magnetic field $\mathbf B$ experiences zero net force and a torque $$\boldsymbol\tau=\mathbf m\times\mathbf B,\qquad \tau=NIAB\sin\theta,$$ where $\theta$ is the angle between $\mathbf m$ (the normal to the loop) and $\mathbf B$. The torque turns $\mathbf m$ towards $\mathbf B$. It is greatest, $NIAB$, when the plane of the loop is parallel to $\mathbf B$ ($\theta=90^\circ$), and zero when the plane is perpendicular to $\mathbf B$: $\theta=0$ is stable equilibrium and $\theta=180^\circ$ is unstable. The result holds for a plane loop of any shape.

Compare Result 1.7: an electric dipole in a uniform electric field feels $\boldsymbol\tau=\mathbf p\times\mathbf E$ and no net force. The current loop is the magnetic counterpart, with $\mathbf m$ in place of $\mathbf p$ and $\mathbf B$ in place of $\mathbf E$. An electric motor keeps a coil turning by reversing its current every half-turn (with a commutator), so that the torque always acts in the same rotational sense.

✏️ Example 1 — Torque on a rectangular coil
A rectangular coil of $50$ turns measures $5.0$ cm $\times$ $4.0$ cm and carries $2.0$ A. It is placed in a uniform field of $0.30$ T with its plane at $30^\circ$ to the field. (a) Find its magnetic moment. (b) Find the torque on it. (c) What is the greatest torque this field can exert on the coil, and in what position? (d) What is the net force on the coil?
(a) Moment $A=(0.050)(0.040)=2.0\times10^{-3}\ \text{m}^2$; $m=NIA=(50)(2.0)(2.0\times10^{-3})=0.20\ \text{A m}^2$.
(b) Torque The plane is at $30^\circ$ to $\mathbf B$, so the normal is at $\theta=60^\circ$: $\tau=mB\sin60^\circ$ $=(0.20)(0.30)(0.866)\approx5.2\times10^{-2}$ N m.
(c) Maximum $\tau_\text{max}=mB=6.0\times10^{-2}$ N m, when the plane of the coil is parallel to $\mathbf B$.
(d) Force Zero — the field is uniform.
Check Interactive 1 uses this coil: at $\theta=60^\circ$ it reads $\tau=0.052$ N m. Watch out: using the angle of the plane ($30^\circ$) in $\sin\theta$ gives $0.030$ N m — the most common slip.
✏️ Example 2 — A circular coil free to turn
A circular coil of $100$ turns and radius $2.0$ cm carries $0.50$ A. It hangs in a uniform horizontal field of $0.10$ T and can turn about a vertical diameter. (a) Find its magnetic moment. (b) Find the torque when its normal makes $0^\circ$, $30^\circ$ and $90^\circ$ with the field. (c) In which position does it come to rest, and why?
(a) Moment $m=NI\pi R^2$ $=(100)(0.50)\pi(0.020)^2$ $\approx6.3\times10^{-2}\ \text{A m}^2$.
(b) Torques $mB=6.28\times10^{-3}$ N m. At $0^\circ$: $0$; at $30^\circ$: $mB\sin30^\circ\approx3.1\times10^{-3}$ N m; at $90^\circ$: $6.3\times10^{-3}$ N m.
(c) Rest It settles with $\mathbf m$ along $\mathbf B$ (plane of the coil perpendicular to the field), the stable position: a small twist gives a torque back towards it. Like a compass needle, the coil lines up with the field.
Interactive 1 — A Coil Turning in a Magnetic Field

The coil of Example 1 seen from above, looking down its axis of rotation: the loop is edge-on, and its two vertical sides cross the page at ⊙ (current up, out of the page) and ⊗ (down). Each of these sides feels a force $NIbB$ at right angles to $\mathbf B$; together they form a couple of arm $a\sin\theta$. Set the angle $\theta$ between $\mathbf m$ and $\mathbf B$ (or drag the coil), and press Play to let it swing.

2.0
angle between m and B—
force on each side (N)—
torque τ (N m)—
m = NIA (A m²)—

The Current Loop as a Magnetic Dipole

In Section 4.2 we found that far along its axis a loop produces $B\approx\dfrac{\mu_0NIR^2}{2x^3}$. With $m=NI\pi R^2$ this is $$B\approx\frac{\mu_0}{4\pi}\,\frac{2m}{x^3},$$ exactly the form of the axial field of an electric dipole, $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{x^3}$, with $\mu_0$ in place of $1/\varepsilon_0$ and $m$ in place of $p$. A longer calculation (not needed here) gives the field in the plane of the loop, far away, as $\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$, opposite to $\mathbf m$ — again the electric-dipole pattern. And the torque in a uniform field has the same form, $\mathbf m\times\mathbf B$ against $\mathbf p\times\mathbf E$. So, far away, a small current loop is indistinguishable from a magnetic dipole.

📐 Result 4.12 — Far field of a current loop (magnetic dipole)

At a distance $r$ large compared with its size, a plane current loop of magnetic moment $\mathbf m$ produces:

on its axis: $\ \mathbf B\approx\dfrac{\mu_0}{4\pi}\,\dfrac{2\mathbf m}{r^3}$; $\quad$ in its plane (equatorial): $\ \mathbf B\approx-\dfrac{\mu_0}{4\pi}\,\dfrac{\mathbf m}{r^3}$.

The pattern of field lines far away is that of an electric dipole, with $\mathbf p\to\mathbf m$ and $\dfrac{1}{\varepsilon_0}\to\mu_0$.

There is one deep difference. An electric dipole is made of two separate charges, and you can pull them apart. A magnetic dipole is a circulating current: no one has ever found an isolated "magnetic charge" (a monopole). Close to a current loop the field lines pass straight through it and close on themselves; there is nowhere for them to start or stop. In Chapter 5 we shall see that a bar magnet is itself a magnetic dipole of the same kind, produced by currents inside atoms.

✏️ Example 3 — A small coil seen from far away
A coil of $50$ turns and radius $3.0$ cm carries $1.0$ A. (a) Find its magnetic moment. (b) Find the field on its axis $30$ cm from the centre using the dipole formula, and compare with the exact Result 4.5. (c) Estimate the field at $30$ cm in the plane of the coil.
(a) Moment $m=NI\pi R^2=(50)(1.0)\pi(0.030)^2\approx0.141\ \text{A m}^2$.
(b) Axis Dipole: $B\approx\dfrac{(10^{-7})(2)(0.141)}{(0.30)^3}\approx1.05\times10^{-6}$ T. Exact: $\dfrac{\mu_0NIR^2}{2(x^2+R^2)^{3/2}}\approx1.03\times10^{-6}$ T. The dipole formula is only $1.5\%$ high at $x=10R$.
(c) Plane Half of the axial value at the same distance: $\approx5.2\times10^{-7}$ T, directed opposite to $\mathbf m$.

The Moving-Coil Galvanometer

A galvanometer detects and measures small currents. In the moving-coil type, a rectangular coil of many turns of fine insulated wire is wound on a light frame and pivoted (or hung on a fine suspension) between the poles of a permanent magnet. Two design features make it a precise instrument:

1. A radial field. The pole pieces are cut concave, and a soft-iron cylinder sits inside the coil. The field in the narrow gap then points along the radii of the cylinder, so whichever way the coil has turned, its plane is always parallel to the field at its sides. The torque is therefore always the maximum value, $\tau=NIAB$, independent of the deflection. (The soft iron also concentrates the field, making it stronger.)

2. A restoring spring. A fine spiral spring (or the twist of the suspension) exerts a restoring torque $k\phi$ proportional to the angle $\phi$ through which the coil has turned; $k$ is the torsional constant of the spring, in N m per radian. A pointer fixed to the coil moves over a scale.

The coil turns until the spring torque balances the magnetic torque: $$k\phi=NIAB\quad\Longrightarrow\quad\phi=\left(\frac{NAB}{k}\right)I.$$ The deflection is proportional to the current, so the scale is uniform (linear) — a direct consequence of the radial field.

📘 Definition 4.6 — Current and voltage sensitivity of a galvanometer
The current sensitivity of a galvanometer is the deflection per unit current, and its voltage sensitivity is the deflection per unit potential difference across it: $$\frac{\phi}{I}=\frac{NAB}{k},\qquad \frac{\phi}{V}=\frac{\phi}{IG}=\frac{NAB}{kG},$$ where $G$ is the resistance of the galvanometer coil. They are measured in rad/A (or divisions per ampere) and rad/V (or divisions per volt).
📐 Result 4.13 — The moving-coil galvanometer

In a radial field $B$, a coil of $N$ turns and area $A$, held by a spring of torsional constant $k$, comes to rest at the deflection $$\phi=\frac{NAB}{k}\,I,$$ proportional to the steady current $I$. Increasing $N$, $A$ or $B$, or using a weaker spring, increases the current sensitivity. Increasing $N$ also increases $G$ (more wire), so it need not increase the voltage sensitivity: doubling the turns of the same wire doubles both $NAB/k$ and $G$, leaving $\phi/V$ unchanged.

✏️ Example 4 — Sensitivity of a galvanometer
A galvanometer coil has $100$ turns, each of area $2.0\ \text{cm}^2$, in a radial field of $0.25$ T. The pointer reaches full scale, a deflection of $90^\circ$, for a current of $2.0$ mA. The coil resistance is $50\ \Omega$. Find (a) the current sensitivity in rad/mA and in degrees per mA, (b) the torsional constant of the spring, (c) the voltage sensitivity, and (d) the current that produces a deflection of $30^\circ$.
(a) Current sens. $\dfrac{\phi}{I}=\dfrac{\pi/2}{2.0\ \text{mA}}\approx0.785$ rad/mA $=45^\circ$ per mA.
(b) Spring $NAB=(100)(2.0\times10^{-4})(0.25)=5.0\times10^{-3}$ N m/A, and $\dfrac{NAB}{k}=785$ rad/A, so $k=\dfrac{5.0\times10^{-3}}{785}\approx6.4\times10^{-6}$ N m/rad — a very weak spring.
(c) Voltage sens. $\dfrac{\phi}{V}=\dfrac{785\ \text{rad/A}}{50\ \Omega}\approx15.7$ rad/V $\approx0.90^\circ$ per mV.
(d) 30° The scale is linear: $I=\dfrac{30^\circ}{45^\circ\ \text{per mA}}\approx0.67$ mA.
Check At equilibrium the two torques match: at full scale $NIAB=(5.0\times10^{-3})(2.0\times10^{-3})$ $=1.0\times10^{-5}$ N m and $k\phi=(6.37\times10^{-6})(1.571)$ $=1.0\times10^{-5}$ N m. ✓ Interactive 2 is built on this galvanometer.

Converting a Galvanometer into an Ammeter or a Voltmeter

A galvanometer alone is a poor meter: full scale is reached at a few milliamperes, and its resistance (tens of ohms) would change the circuit it is put in. Both problems are solved by adding one resistor. (We use the series and parallel rules for resistors from Class 10, and Kirchhoff's rules from Chapter 3.)

Ammeter. An ammeter goes in series in the circuit, so it should have a very small resistance. Connect a small resistance $S$, called a shunt, in parallel with the galvanometer. Of the total current $I$, only $I_g$ goes through the coil; the rest, $I-I_g$, bypasses it through the shunt. The two branches have the same potential difference: $$I_gG=(I-I_g)S\quad\Longrightarrow\quad S=\frac{I_gG}{I-I_g}.$$ If $I_g$ is the full-scale current of the galvanometer, the combination reads full scale for a total current $I$. Its resistance, $\dfrac{GS}{G+S}$, is less than $S$ — very small.

Voltmeter. A voltmeter goes in parallel with the component whose potential difference it measures, so it should have a very large resistance, drawing almost no current. Connect a large resistance $R$ in series with the galvanometer. A potential difference $V$ across the combination drives the current $I_g=V/(G+R)$, so for full scale at $V$: $$V=I_g(G+R)\quad\Longrightarrow\quad R=\frac{V}{I_g}-G.$$

📐 Result 4.14 — Conversion of a galvanometer

A galvanometer of resistance $G$ that gives full-scale deflection for a current $I_g$ becomes:

an ammeter of range $0$–$I$ with a shunt $\ S=\dfrac{I_gG}{I-I_g}\ $ in parallel (meter resistance $\dfrac{GS}{G+S}$, very small);

a voltmeter of range $0$–$V$ with a resistance $\ R=\dfrac{V}{I_g}-G\ $ in series (meter resistance $G+R=V/I_g$, very large).

An ideal ammeter has zero resistance and an ideal voltmeter infinite resistance; a real meter disturbs the circuit slightly.

✏️ Example 5 — Making an ammeter and a voltmeter
The galvanometer of Example 4 ($G=50\ \Omega$, full scale at $I_g=2.0$ mA) is to be converted into (a) an ammeter reading up to $5.0$ A and (b) a voltmeter reading up to $10$ V. Find the resistor needed in each case, how it is connected, and the resistance of the finished meter.
(a) Shunt $$S=\frac{I_gG}{I-I_g}=\frac{(2.0\times10^{-3})(50)}{5.0-0.0020}\approx2.0\times10^{-2}\ \Omega,$$ in parallel. Meter resistance $\dfrac{GS}{G+S}\approx0.020\ \Omega$ — the shunt dominates.
(b) Series R $R=\dfrac{V}{I_g}-G=\dfrac{10}{2.0\times10^{-3}}-50=4950\ \Omega$, in series. Meter resistance $5000\ \Omega$.
Check (a) At full scale the shunt carries $4.998$ A, so the p.d. across it is $(4.998)(0.0200)=0.100$ V, and the coil carries $0.100/50=2.0$ mA. ✓ (b) $(2.0\ \text{mA})(5000\ \Omega)=10$ V. ✓ Interactive 2 gives the same values.
✏️ Example 6 — Meters disturb the circuit
(a) A $6.0$ V cell of negligible internal resistance drives a current through a $3.0\ \Omega$ resistor. What does an ammeter in series read if its resistance is $0.020\ \Omega$? If it is $1.0\ \Omega$? (b) Two $1000\ \Omega$ resistors are in series across the same cell. A voltmeter of resistance $5000\ \Omega$ is connected across one of them. What does it read, and what is the true p.d. without the meter?
(a) Ammeter Without a meter, $I=6.0/3.0=2.0$ A. With $0.020\ \Omega$: $I=\dfrac{6.0}{3.02}\approx1.99$ A (an error of $0.7\%$). With $1.0\ \Omega$: $I=\dfrac{6.0}{4.0}=1.5$ A — the meter itself has cut the current by a quarter.
(b) Voltmeter Without the meter each resistor has $3.0$ V. With it, the measured resistor is in parallel with $5000\ \Omega$: $\dfrac{(1000)(5000)}{6000}\approx833\ \Omega$. The p.d. across it is $6.0\times\dfrac{833}{1833}\approx2.7$ V — the meter reads $2.7$ V instead of $3.0$ V.
Lesson An ammeter must have a resistance much smaller, and a voltmeter much larger, than the resistances in the circuit. That is why the shunt is small and the series resistor large.
Interactive 2 — Galvanometer, Ammeter and Voltmeter

The galvanometer of Example 4 ($100$ turns, $2.0\ \text{cm}^2$, $B=0.25$ T radial, $G=50\ \Omega$, full scale $2.0$ mA). Left: the coil in the radial field between curved pole pieces — the magnetic torque $NI_gAB$ is balanced by the spring torque $k\phi$. Right: the dial and the instrument's circuit. Choose what to build, set its range, and slide the quantity being measured.

——
current in the coil—
deflection φ—
extra resistor—
meter resistance—
⚠️ Common mistakes with current loops and meters

1. Using the angle of the plane. In $\tau=NIAB\sin\theta$, $\theta$ is the angle between the normal ($\mathbf m$) and $\mathbf B$. A coil whose plane is parallel to $\mathbf B$ has $\theta=90^\circ$ and the largest torque.

2. Expecting a net force. In a uniform field a loop feels only a torque. A net force needs a non-uniform field.

3. Shunt in series, or series resistor in parallel. Ammeter: small shunt in parallel. Voltmeter: large resistor in series.

4. Forgetting $I_g$ in the shunt formula. $S=\dfrac{I_gG}{I-I_g}$, not $\dfrac{I_gG}{I}$ — the difference matters when the range is only a few times $I_g$.

5. "More turns means a more sensitive voltmeter." More turns of the same wire raise the current sensitivity but also the resistance, so the voltage sensitivity is unchanged.

Exercises

I
Torque on current loops.
  • 1 A single square loop of side $10$ cm carries $5.0$ A in a uniform field of $0.40$ T, with its plane parallel to the field. Find the torque on it.
  • 2 The loop of part 1 is turned so that its plane is perpendicular to the field. What is the torque now? Is this position stable?
  • 3 A coil of $200$ turns and area $4.0\times10^{-3}\ \text{m}^2$ carries $0.50$ A. Its normal makes $60^\circ$ with a field of $0.25$ T. Find the torque.
  • 4 Find the magnetic moment of a circular coil of $50$ turns and radius $4.0$ cm carrying $2.0$ A.
  • 5 A loop has $\mathbf m=(0.30\,\hat{\mathbf i}+0.40\,\hat{\mathbf j})\ \text{A m}^2$ in a field $\mathbf B=0.50\,\hat{\mathbf k}$ T. Find the torque vector and its magnitude.
  • 6 A wire of length $L$ carrying a current $I$ is bent into either a circle or a square. Which has the larger magnetic moment, and by what factor?
  • 7 A coil of magnetic moment $0.40\ \text{A m}^2$ experiences a maximum torque of $0.12$ N m in a uniform field. Find the field.
  • 8 Explain why $\theta=0$ is a stable and $\theta=180^\circ$ an unstable equilibrium position for a current loop in a uniform field.
1 Plane parallel to $\mathbf B$ means $\theta=90^\circ$: $\tau=IAB=(5.0)(0.010)(0.40)=2.0\times10^{-2}$ N m.
2 Plane perpendicular to $\mathbf B$: $\mathbf m\parallel\mathbf B$ (or antiparallel), $\tau=0$. It is stable if $\mathbf m$ points along $\mathbf B$, unstable if opposite.
3 $\tau=NIAB\sin60^\circ=(200)(0.50)(4.0\times10^{-3})(0.25)(0.866)\approx8.7\times10^{-2}$ N m.
4 $m=NI\pi R^2=(50)(2.0)\pi(0.040)^2\approx0.50\ \text{A m}^2$.
5 $\boldsymbol\tau=\mathbf m\times\mathbf B=0.50\left[0.30\,(\hat{\mathbf i}\times\hat{\mathbf k})+0.40\,(\hat{\mathbf j}\times\hat{\mathbf k})\right]$ $=(0.20\,\hat{\mathbf i}-0.15\,\hat{\mathbf j})$ N m; magnitude $0.25$ N m (as expected, $\mathbf m\perp\mathbf B$ and $mB=0.50\times0.50$).
6 Circle: $A=\dfrac{L^2}{4\pi}$; square: $A=\dfrac{L^2}{16}$. The circle's moment is larger by $\dfrac{16}{4\pi}=\dfrac4\pi\approx1.27$ — a circle encloses the most area for a given perimeter.
7 $B=\dfrac{\tau_\text{max}}{m}=\dfrac{0.12}{0.40}=0.30$ T.
8 Near $\theta=0$ the torque $mB\sin\theta$ always acts to reduce $\theta$, so a small twist is undone. Near $180^\circ$ the torque is again zero, but a small twist produces a torque that increases the twist further, turning the loop right round.
II
Current loops as magnetic dipoles. $\mu_0/4\pi=10^{-7}$ T m/A.
  • 1 A small coil has $m=0.50\ \text{A m}^2$. Find the field $50$ cm from it (a) on its axis and (b) in its plane.
  • 2 At the same large distance, what is the ratio of the axial field to the equatorial field of a magnetic dipole?
  • 3 Show that the units $\text{A m}^2$ and $\text{J/T}$ are the same.
  • 4 A coil of $10$ turns and radius $1.0$ cm carries $1.0$ A. Find its magnetic moment and its field on the axis $20$ cm away.
  • 5 List three ways in which a small current loop behaves like an electric dipole, and one way in which it differs.
  • 6 A loop in the plane of the page carries an anticlockwise current. Which way does $\mathbf m$ point? Which way does its field point at a distant point in the plane of the page?
1 (a) $B=\dfrac{(10^{-7})(2)(0.50)}{(0.50)^3}=8.0\times10^{-7}$ T, along $\mathbf m$. (b) Half: $4.0\times10^{-7}$ T, opposite to $\mathbf m$.
2 $2:1$, just as for an electric dipole.
3 From $\tau=mB$: $m=\tau/B$ has unit $\text{N m/T}=\text{J/T}$. Also $1\ \text{T}=1\ \text{N A}^{-1}\text{m}^{-1}$, so $\text{J/T}=\text{N m}\cdot\text{A m/N}=\text{A m}^2$. ✓
4 $m=(10)(1.0)\pi(0.010)^2\approx3.1\times10^{-3}\ \text{A m}^2$. $B\approx\dfrac{(10^{-7})(2)(3.14\times10^{-3})}{(0.20)^3}\approx7.9\times10^{-8}$ T.
5 Alike: far field falls as $1/r^3$; axial field twice the equatorial; torque $\mathbf m\times\mathbf B$ like $\mathbf p\times\mathbf E$; no net force in a uniform field. Different: the magnetic dipole cannot be split into separate "magnetic charges" — its field lines are closed loops passing through it.
6 $\mathbf m$ points out of the page (right-hand thumb rule). A distant point in the plane of the loop is on its equator, where the field is opposite to $\mathbf m$: into the page.
III
The moving-coil galvanometer.
  • 1 Why are the pole pieces of a moving-coil galvanometer made concave, with a soft-iron cylinder inside the coil?
  • 2 A galvanometer coil has $150$ turns of area $3.0\ \text{cm}^2$ in a radial field of $0.20$ T, and the spring constant is $4.5\times10^{-6}$ N m/rad. Find the current sensitivity and the deflection for $0.10$ mA.
  • 3 The coil of part 2 has resistance $40\ \Omega$. Find the voltage sensitivity and the deflection produced by $1.0$ mV.
  • 4 The number of turns of the coil of part 2 is doubled using the same wire. What happens to the current sensitivity and to the voltage sensitivity?
  • 5 For a galvanometer with $NAB=6.0\times10^{-4}$ N m/A, what spring constant gives full-scale deflection ($90^\circ$) at $50\ \mu\text{A}$?
  • 6 Give two reasons why the scale of a moving-coil galvanometer is uniform.
  • 7 Why can a galvanometer not be used directly to measure the current in a circuit carrying a few amperes?
1 To make the field radial, so the plane of the coil is always parallel to $\mathbf B$ and the torque $NIAB$ does not depend on the deflection; the soft iron also strengthens the field in the gap.
2 $\dfrac{\phi}{I}=\dfrac{NAB}{k}=\dfrac{(150)(3.0\times10^{-4})(0.20)}{4.5\times10^{-6}}=2.0\times10^{3}$ rad/A. For $0.10$ mA: $\phi=0.20$ rad $\approx11^\circ$.
3 $\dfrac{\phi}{V}=\dfrac{2000}{40}=50$ rad/V; for $1.0$ mV, $\phi=0.050$ rad $\approx2.9^\circ$.
4 $NAB/k$ doubles, so the current sensitivity doubles. $G$ doubles too (twice the wire), so the voltage sensitivity $NAB/(kG)$ is unchanged.
5 $k=\dfrac{NAB\,I}{\phi}=\dfrac{(6.0\times10^{-4})(5.0\times10^{-5})}{\pi/2}\approx1.9\times10^{-8}$ N m/rad.
6 The radial field keeps the magnetic torque equal to $NIAB$ at every deflection, and the spring torque $k\phi$ is proportional to $\phi$; so $\phi\propto I$.
7 It reaches full scale at a few milliamperes (larger currents would damage the fine coil and spring), and its resistance would add noticeably to the circuit. A shunt is needed (an ammeter).
IV
Ammeters and voltmeters.
  • 1 A galvanometer of resistance $100\ \Omega$ gives full scale for $1.0$ mA. Find the shunt needed to make it an ammeter of range $0$–$1.0$ A.
  • 2 Find the resistance needed to convert the same galvanometer into a voltmeter of range $0$–$5.0$ V, and say how it is connected.
  • 3 Find the resistance of the ammeter of part 1.
  • 4 A galvanometer of resistance $25\ \Omega$ gives full scale for $4.0$ mA. A shunt of $0.10\ \Omega$ is connected across it. What is the range of the resulting ammeter?
  • 5 A voltmeter has a range of $3.0$ V and a resistance of $3000\ \Omega$. What must be added to extend its range to $15$ V?
  • 6 Explain why an ammeter is connected in series and should have a low resistance, while a voltmeter is connected in parallel and should have a high resistance.
  • 7 Two $2.0\ \text{k}\Omega$ resistors in series are connected to a $12$ V supply of negligible internal resistance. A voltmeter of resistance $2.0\ \text{k}\Omega$ is connected across one of them. What does it read?
  • 8 An ammeter of resistance $0.50\ \Omega$ is placed in series with a $2.5\ \Omega$ resistor across an ideal $6.0$ V cell. What does it read, and what was the current before it was inserted?
1 $S=\dfrac{I_gG}{I-I_g}=\dfrac{(1.0\times10^{-3})(100)}{0.999}\approx0.10\ \Omega$ (more precisely $0.1001\ \Omega$), in parallel.
2 $R=\dfrac{V}{I_g}-G=\dfrac{5.0}{1.0\times10^{-3}}-100=4900\ \Omega$, in series.
3 $\dfrac{GS}{G+S}=\dfrac{(100)(0.1001)}{100.1}\approx0.10\ \Omega$.
4 $I=I_g\dfrac{G+S}{S}=(4.0\times10^{-3})\dfrac{25.1}{0.10}\approx1.0$ A.
5 Full scale needs $I_g=\dfrac{3.0}{3000}=1.0$ mA; for $15$ V the total resistance must be $\dfrac{15}{1.0\times10^{-3}}=15\ \text{k}\Omega$. Add $12\ \text{k}\Omega$ in series.
6 An ammeter must carry the current it measures, so it goes in series; a large resistance would reduce that current. A voltmeter compares the potentials at two points, so it goes across them; it must draw negligible current, otherwise it changes the p.d. it is measuring.
7 The measured resistor in parallel with the meter: $\dfrac{(2.0)(2.0)}{4.0}=1.0\ \text{k}\Omega$. The reading is $12\times\dfrac{1.0}{3.0}=4.0$ V, instead of the true $6.0$ V.
8 With the ammeter: $\dfrac{6.0}{3.0}=2.0$ A. Without it: $\dfrac{6.0}{2.5}=2.4$ A. The meter has reduced the current by about $17\%$.
🎯 Exam corner — torque on loops and moving-coil meters BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Derive an expression for the torque on a rectangular coil of $N$ turns and area $A$, carrying a current $I$, placed in a uniform magnetic field $\mathbf B$ with its normal at angle $\theta$ to $\mathbf B$. Find the torque on a coil of $100$ turns measuring $4.0$ cm $\times$ $5.0$ cm, carrying $0.20$ A, with its plane parallel to a field of $0.50$ T.

Forces Sides of length $a$ (in the plane containing $\mathbf B$ and the normal): forces equal, opposite and collinear — no torque. Sides of length $b$ (perpendicular to $\mathbf B$): forces $NIbB$, equal and opposite, along parallel lines a distance $a\sin\theta$ apart (diagram).
Torque Couple: $\tau=NIbB\cdot a\sin\theta=NIAB\sin\theta$, i.e. $\boldsymbol\tau=\mathbf m\times\mathbf B$ with $\mathbf m=NIA\,\hat{\mathbf n}$.
Numbers Plane parallel to $\mathbf B$: $\theta=90^\circ$. $\tau=(100)(0.20)(2.0\times10^{-3})(0.50)=2.0\times10^{-2}$ N m. Exam tip: state clearly that $\theta$ is measured from the normal.

E2. Board · 2 marks Why is the magnetic field in a moving-coil galvanometer made radial? Define its current sensitivity. How is the voltage sensitivity affected if the number of turns is doubled using the same wire?

Radial field The plane of the coil is then always parallel to $\mathbf B$, so the deflecting torque is $NIAB$ at every position; balanced by $k\phi$, it gives $\phi\propto I$ — a linear scale.
Sensitivity Current sensitivity $=\phi/I=NAB/k$, the deflection per unit current.
Doubling N $NAB/k$ doubles but $G$ also doubles, so the voltage sensitivity $NAB/(kG)$ is unchanged. Exam tip: this "trick" question appears often — always ask what happens to $G$.

E3. JEE Main A galvanometer of resistance $99\ \Omega$ gives full-scale deflection for $10$ mA. To use it as an ammeter of range $0$–$1.0$ A, one must connect (A) $0.10\ \Omega$ in parallel (B) $1.0\ \Omega$ in parallel (C) $9.9\ \Omega$ in parallel (D) $99\ \Omega$ in series.

Formula $S=\dfrac{I_gG}{I-I_g}=\dfrac{(0.010)(99)}{1.0-0.010}=\dfrac{0.99}{0.99}=1.0\ \Omega$, in parallel.
Answer (B). Exam tip: (A) is $I_gG/I\approx0.99\ \Omega$ rounded carelessly — here keeping $I-I_g$ gives a neat exact answer, a hint the examiner intended it.

E4. NEET A galvanometer of resistance $20\ \Omega$ gives full-scale deflection for $5.0$ mA. The resistance required in series to convert it into a voltmeter of range $0$–$10$ V is (A) $1.98\ \Omega$ (B) $1980\ \Omega$ (C) $2000\ \Omega$ (D) $2020\ \Omega$.

Formula $R=\dfrac{V}{I_g}-G=\dfrac{10}{5.0\times10^{-3}}-20=2000-20=1980\ \Omega$.
Answer (B). Exam tip: (C) is the total resistance of the voltmeter, not the resistor to be added.

E5. JEE Advanced (One or more correct.) A circular loop of radius $R$ carrying a current $I$ is placed in a uniform magnetic field $\mathbf B$ that lies in the plane of the loop. Which statements are correct? (A) The net force on the loop is zero. (B) The torque on the loop has magnitude $I\pi R^2B$. (C) The torque tends to turn the loop so that its magnetic moment lines up with $\mathbf B$. (D) The loop is in stable equilibrium in this position.

Options (A) Uniform field: true. (B) $\mathbf B$ in the plane means $\mathbf m\perp\mathbf B$: $\tau=mB=I\pi R^2B$: true. (C) $\boldsymbol\tau=\mathbf m\times\mathbf B$ turns $\mathbf m$ towards $\mathbf B$: true. (D) The torque is not zero here, so this is not an equilibrium at all: false.
Answer (A), (B), (C). Exam tip: equilibrium (stable or unstable) needs zero torque: $\mathbf m$ parallel or antiparallel to $\mathbf B$.

E6. Board · Case study A student has a galvanometer of resistance $40\ \Omega$ that gives full-scale deflection for $2.5$ mA. (i) (2 marks) Show how she can convert it into an ammeter of range $0$–$2.5$ A; find the resistor needed and the resistance of the ammeter. (ii) (1 mark) Find the resistor needed instead for a voltmeter of range $0$–$25$ V. (iii) (1 mark) Which meter has the higher resistance, and why is that desirable?

(i) Shunt in parallel: $S=\dfrac{(2.5\times10^{-3})(40)}{2.5-0.0025}\approx4.0\times10^{-2}\ \Omega$. Ammeter resistance $\dfrac{GS}{G+S}\approx0.040\ \Omega$.
(ii) In series: $R=\dfrac{25}{2.5\times10^{-3}}-40=9960\ \Omega$.
(iii) The voltmeter ($10\ \text{k}\Omega$) has a far higher resistance than the ammeter ($0.04\ \Omega$). A voltmeter in parallel must draw almost no current; an ammeter in series must hardly reduce the current. Exam tip: draw both circuit diagrams — they carry a mark.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How would you design the analogue current and voltage meters on the panel of a small solar charge controller, and why are real shunts made of special alloys rather than copper?

Why it matters. Off-grid solar systems power homes, clinics and schools in rural India, Africa and South America, and their charge controllers display the charging current and battery voltage. Even in digital meters the current is measured by the small voltage across a precision shunt resistor — the same idea as Result 4.14. A shunt that warms up and changes its resistance gives wrong readings.

What to do.

  1. You have a moving-coil movement with $G=60\ \Omega$ and full scale at $1.0$ mA. Find the shunt for a $0$–$10$ A ammeter and the series resistor for a $0$–$30$ V voltmeter.
  2. Find the power dissipated in the shunt and in the series resistor at full scale. Which needs the larger power rating?
  3. Suppose the shunt were made of copper, whose resistance rises by about $0.43\%$ per kelvin (Chapter 3), and it warms by $20$ K at full current while the coil stays cool. By how much, and in which direction, is the ammeter reading wrong? Repeat for manganin (about $0.002\%$ per kelvin).
  4. Research one other reason meter makers use manganin or constantan for shunts, and how four-terminal shunts avoid errors from contact resistance.

Tools: Calculator; data on resistivity and temperature coefficients of alloys.

Hand in: A labelled circuit diagram of each meter with component values and power ratings, your error estimates, and a paragraph recommending a shunt material.

Going further. A voltmeter's "ohms per volt" figure is $1/I_g$. Find this figure for your meter and for a typical digital multimeter (input resistance about $10\ \text{M}\Omega$), and explain which disturbs a high-resistance circuit less.

Instructor note: $S=\dfrac{(10^{-3})(60)}{9.999}\approx6.0\times10^{-3}\ \Omega$; $R=\dfrac{30}{10^{-3}}-60=29\,940\ \Omega$. Power: shunt $I^2S\approx(10)^2(0.006)=0.60$ W; series resistor $\approx(10^{-3})^2(29\,940)\approx0.030$ W — the shunt needs the larger rating. Copper shunt $+20$ K: $S$ rises $8.6\%$, so a larger fraction of current goes through the coil: the meter reads about $8.6\%$ high (many students guess "low" — a good discussion point; the coil current is $IS/(G+S)\approx IS/G\propto S$). Manganin: about $0.04\%$. Four-terminal (Kelvin) connection measures the voltage across the shunt element itself, excluding lead and contact resistances. Ohms per volt: $1000\ \Omega/\text{V}$; at $30$ V range, $30\ \text{k}\Omega$ versus $10\ \text{M}\Omega$ for the DMM.

🧪 Self-check quiz

0/5 correct

Q1. The torque on a current loop in a uniform magnetic field is greatest when

Q2. The SI unit of magnetic dipole moment is

Q3. In a moving-coil galvanometer the magnetic field is made radial so that

Q4. To convert a galvanometer into an ammeter, one connects

Q5. The current sensitivity of a moving-coil galvanometer can be increased by

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