Take a breath before you start this one. Moving Charges and Magnetism has a reputation in Class 12 for being the chapter where directions suddenly stop behaving. Up to now, forces have pointed along the line joining two things — push or pull, and that was that. Here, for the first time, a force points sideways. A charge zips forward, a field runs across it, and the push comes out perpendicular to both. Students who quietly panic at this point are not bad at physics; they are simply meeting a new habit of thought for the first time. Give it three days and it becomes second nature.
Here is the promise of this page. We will build one single mental picture — the Traffic Cop Grip of your right hand — and then use that same picture, unchanged, for every direction question in the chapter: the force on a lone moving charge, the force on a current-carrying wire, the pull between two parallel wires, and the twist on a current loop. One picture, four uses. Alongside it you get Biot-Savart law solved examples worked line by line, a tidy moving charges and magnetism formula list class 12 you can revise in four minutes, a full set of force between two parallel current carrying conductors numericals, and a rundown of the moving charges and magnetism class 12 important questions that examiners keep coming back to.
If you have already worked through the Electric Charges and Fields chapter, you have exactly the right background. Everything there was about charges sitting still. This chapter is what happens the moment they start moving. Same characters, new plot.
Meet Your Tutor
Magnetism questions become manageable when the direction is decided before the magnitude. I will help you use the right-hand rules, Lorentz force, Biot-Savart and Ampere laws in a fixed order, then check every numerical for vector direction and units.
What You’ll Learn
- Magnetic Field and Oersted’s Experiment: Where It All Starts
- Biot-Savart Law Solved Examples: The Field of a Current Element
- Biot-Savart Law Applied to a Current Carrying Circular Loop
- Ampere’s Circuital Law and the Infinitely Long Straight Wire
- The Straight Solenoid — Qualitative Treatment Only
- Force on a Moving Charge in Uniform Magnetic and Electric Fields
- Force on a Current-Carrying Conductor in a Uniform Magnetic Field
- Force Between Two Parallel Current Carrying Conductors Numericals and the Definition of the Ampere
- Torque Experienced by a Current Loop in a Uniform Magnetic Field
- Current Loop as a Magnetic Dipole and Its Magnetic Dipole Moment
- Moving Coil Galvanometer and Its Current Sensitivity
- Conversion of a Galvanometer into an Ammeter and a Voltmeter
- Moving Charges and Magnetism Formula List Class 12
- Moving Charges and Magnetism Class 12 Important Questions
- What’s NOT in Your 2026-27 Syllabus
Your Game Plan
- Day 1 — learn the grip. Do nothing but the Traffic Cop Grip. Point your right hand at the ceiling, the wall, your desk. Get the directions into your muscles before you touch a formula.
- Day 2 — sources of field. Biot-Savart law, the circular loop, Ampere’s law for a straight wire, and a qualitative look at the solenoid. These are the “where does B come from” topics.
- Day 3 — effects of field. Force on a charge, force on a wire, force between two wires, torque on a loop. These are the “what does B do” topics.
- Day 4 — the instrument. Moving coil galvanometer, current sensitivity, and the two conversions. This block is quietly one of the most repeated in board papers.
- Day 5 — the worksheet. Close the notes. Attempt all ten questions cold. Only then open the answers.
Study Notes
Magnetic Field and Oersted’s Experiment: Where It All Starts
Start from zero. A magnet has a region around it where another magnet feels a push or a twist. We call that region a magnetic field and give it the symbol B. It is a vector: it has a size and a direction at every point in space. Its SI unit is the tesla (T), and one tesla is a genuinely large field — the Earth’s own field is around 5 × 10−5 T, so a 1 T laboratory magnet is roughly twenty thousand times stronger than the planet you are standing on.
For centuries, electricity and magnetism were thought to be two separate subjects. Then in 1820 Hans Christian Oersted was demonstrating a circuit to a class, and a compass needle lying on the bench swung sideways the moment he closed the switch. Open the switch, the needle drifted back. That accident is the hinge of this entire chapter.
Read carefully what the needle did. It did not point along the wire, and it did not point at the wire. It settled perpendicular to the wire, tangent to a circle drawn around it. That is your very first clue that magnetism is a sideways business. The magnetic field of a straight current-carrying wire wraps around the wire in closed circles, like ripples in a pond seen from above — except that here the “ripples” are the field itself and they never stop.
To find which way the circles run, use the right-hand thumb rule: grip the wire with your right hand so that your thumb points along the conventional current. Your curled fingers then sweep the direction of B. Try it now with a pencil. If the current runs upwards out of your desk towards you, the field circles anticlockwise as you look down the wire towards yourself.
Biot-Savart Law Solved Examples: The Field of a Current Element
Oersted told us a field exists. The Biot-Savart law tells us exactly how big it is. The idea is beautifully simple once you see the trick: we cannot easily write down the field of a whole complicated circuit, so we chop the circuit into tiny pieces, work out the field of one tiny piece, and add up.
Take a tiny length dl of wire carrying current I. This little arrow of current is called a current element, written I dl. Look at a point P a distance r away from it, with θ being the angle between the direction of the current element and the line joining it to P. Then the small piece of field that this element contributes at P is
dB = (μ0 / 4π) · I dl sinθ / r2
In vector form, dB = (μ0/4π) · I (dl × r̂) / r2, where r̂ is the unit vector from the element towards P. The constant μ0 is the permeability of free space, and it has the exact value
μ0 = 4π × 10−7 T·m/A, so μ0/4π = 1 × 10−7 T·m/A
That second form is the one you actually use. Whenever a Biot-Savart question appears, mentally replace μ0/4π by 10−7 and the arithmetic collapses into something you can do without a calculator.
Step 1 — convert. dl = 1 mm = 1 × 10−3 m, r = 0.5 m, sin 30° = 0.5.
Step 2 — substitute. dB = 10−7 × (10 × 1 × 10−3 × 0.5) / (0.5)2
Step 3 — simplify. Numerator inside brackets = 5 × 10−3; divide by r2 = 0.25 to get 2 × 10−2; multiply by 10−7.
dB = 2 × 10−9 T = 2 nT.
Tiny — and it should be. One millimetre of wire is not going to move a compass. It is the sum over millions of such elements that gives a measurable field.
The next example shows the pattern you will use again and again: for a circular arc, every element sits at the same distance R from the centre and is always perpendicular to the line joining it to the centre, so sinθ = 1 everywhere. That makes the sum trivial.
Step 1 — the general arc result. For a full circle the total length of wire is 2πR and every element contributes with sinθ = 1, giving B = μ0I/2R. An arc subtending angle φ at the centre gives the same fraction of that: B = (φ/2π) × μ0I/2R.
Step 2 — semicircle means φ = π, which is half a circle, so B = μ0I / 4R.
Step 3 — substitute. B = (4π × 10−7 × 4) / (4 × 0.05) = (5.027 × 10−6) / 0.2
B = 2.51 × 10−5 T, directed perpendicular to the plane of the semicircle.
The two straight lead-in wires, if they lie along a diameter pointing at the centre, contribute nothing, because for them θ = 0° and sinθ = 0. That is the Biot-Savart law doing you a favour.
Biot-Savart Law Applied to a Current Carrying Circular Loop
This is the one application of Biot-Savart that the syllabus names explicitly, so learn it properly. There are two results to hold: the field at the centre, and the field on the axis.
At the centre. Every element of the loop is at distance R from the centre, and every element is perpendicular to its own radius, so sinθ = 1 throughout. Adding all the dl around the loop gives the circumference 2πR:
B = μ0 I / 2R (single turn) → B = μ0 N I / 2R (N tightly wound turns)
On the axis. Move a distance x along the axis from the centre. Now each element sits a slanted distance √(R2 + x2) away. When you add up all the little dB vectors, the components perpendicular to the axis cancel in pairs from opposite sides of the loop, and only the axial components survive. The result is
B = μ0 N I R2 / 2(R2 + x2)3/2
Why it works. The cancellation is the whole story. Pick any element on the loop; there is an element diametrically opposite it. Their two dB vectors are tilted mirror images. The sideways parts point in opposite directions and vanish; the along-the-axis parts point the same way and add. Nature has done the vector algebra for you by symmetry. Notice, too, that setting x = 0 in the axial formula returns μ0NI/2R exactly — a good self-check that you have written it correctly.
Step 1. N = 50, I = 2 A, R = 0.10 m.
Step 2. B = μ0NI / 2R = (4π × 10−7 × 50 × 2) / (2 × 0.10)
Step 3. Numerator = 4π × 10−7 × 100 = 4π × 10−5. Divide by 0.20 to get 2π × 10−4.
B = 2π × 10−4 T = 6.28 × 10−4 T, perpendicular to the plane of the coil.
Sanity check: that is about twelve times the Earth’s field, which is exactly why a small coil like this will visibly swing a compass needle.
Step 1. R = 0.04 m, x = 0.03 m, I = 3 A, N = 1.
Step 2 — do the awkward bracket first. R2 + x2 = 0.0016 + 0.0009 = 0.0025, so √(R2+x2) = 0.05 m and (R2+x2)3/2 = (0.05)3 = 1.25 × 10−4.
Step 3. B = (4π × 10−7 × 3 × 0.0016) / (2 × 1.25 × 10−4) = (6.032 × 10−9) / (2.5 × 10−4)
B = 2.41 × 10−5 T, along the axis.
Compare with the centre value for the same loop, μ0I/2R = 4.71 × 10−5 T. Moving out just 3 cm has roughly halved the field. Examiners love that comparison.
Ampere’s Circuital Law and the Infinitely Long Straight Wire
Biot-Savart always works, but it can be laborious. Ampere’s circuital law is the shortcut you reach for whenever a problem has enough symmetry. It plays exactly the role that Gauss’s law played in electrostatics — if you were comfortable choosing a Gaussian surface in the Electric Charges and Fields chapter, you already know the rhythm of this argument.
The law says: draw any closed loop in space (an Amperian loop). Walk around it, and at each step multiply the component of B along your direction of travel by the length of that step. Add all those products up. The total equals μ0 times the current threading through the loop:
∮ B · dl = μ0 Ienclosed
Applying it to an infinitely long straight wire. Here is the reasoning, step by step, exactly as you should write it in a board answer.
- Symmetry. The wire looks the same from every direction around it and from every point along it. So B can only depend on the perpendicular distance r, and it must be tangent to circles centred on the wire (Oersted showed us this).
- Choose the loop. Take a circle of radius r centred on the wire, lying in a plane perpendicular to it. On this circle, B has the same magnitude everywhere and is everywhere parallel to dl.
- Do the sum. Because B is constant and parallel to dl, the sum becomes B × (circumference) = B · 2πr.
- Equate. B · 2πr = μ0I.
B = μ0 I / 2πr
Look at what that tells you. The field of a straight wire falls off as 1/r, not 1/r2. Double your distance and the field halves — it does not quarter. That single fact catches out a lot of otherwise well-prepared students, because electrostatics trains you to expect inverse squares everywhere.
B = 2 × 10−7 × 15 / 0.06 = 2 × 10−7 × 250
B = 5 × 10−5 T, tangent to the circle of radius 6 cm around the wire.
That is almost exactly the Earth’s magnetic field. Hold a compass 6 cm from such a wire and it will swing about 45° — a lovely home experiment if you have a torch battery and some wire.
Rearrange first, substitute second. r = μ0I / 2πB = 2 × 10−7 × 8 / (4 × 10−5)
= 1.6 × 10−6 / 4 × 10−5
r = 0.04 m = 4 cm.
Step 1 — distances. The midpoint is 5 cm from each wire.
Step 2 — magnitudes.
B1 = 2 × 10−7 × 6 / 0.05 = 2.4 × 10−5 T
B2 = 2 × 10−7 × 10 / 0.05 = 4.0 × 10−5 T
Step 3 — directions. Apply the right-hand thumb rule to each wire separately. Because the currents are antiparallel, at the midpoint both fields point the same way, so they add.
Bnet = 6.4 × 10−5 T.
Change the question so the currents are parallel and the answer flips to a subtraction: 4.0 − 2.4 = 1.6 × 10−5 T. Always settle direction before you decide whether to add or subtract.
The Straight Solenoid — Qualitative Treatment Only
Read this heading twice, because it will save you hours. For the 2026-27 syllabus the straight solenoid is prescribed as qualitative treatment only. You are expected to describe it, sketch its field, and know the result — you are not expected to derive it by applying Ampere’s law to a rectangular loop. Do not spend a week perfecting a derivation the paper will not ask for.
So, qualitatively: a solenoid is a long coil of insulated wire wound in a tight helix, like a spring. Each turn is essentially a circular loop, and each loop throws its field along the common axis. Inside the solenoid all those contributions line up and reinforce each other, giving a field that is strong, uniform, and parallel to the axis. Outside, the field spreads out into the surrounding space and becomes very weak — for a long solenoid it is treated as negligible.
The upshot is that a current-carrying solenoid behaves like a bar magnet: one end acts as a north pole, the other as a south pole, and which is which is settled by curling your right-hand fingers along the current so the thumb points to the north end. Slip a soft iron rod inside and you have an electromagnet whose strength you can switch on and off — the basis of relays, doorbells and scrapyard cranes.
Force on a Moving Charge in Uniform Magnetic and Electric Fields
Now for the heart of the chapter. Send a charge q flying with velocity v through a magnetic field B. The force it feels is
F = q(v × B) → magnitude F = qvB sinθ
Everything strange about magnetism lives inside that cross product, so let us unpack it slowly using the Traffic Cop Grip shown above. Hold up your right hand as if stopping traffic. Your thumb points along the motion, v. Your fingers point along the field, B. Your palm then faces the direction it would push a car — and that is the direction of the force on a positive charge. For a negative charge, everything is the same except the answer is reversed at the very end.
The reason to give this grip a name and use it stubbornly is that you will meet the same cross product three more times in this chapter. Learn one gesture, not four rules.
What follows from a force that is always sideways and always the same size? A circle. If v is perpendicular to a uniform B, the magnetic force supplies exactly the centripetal force, so qvB = mv2/r, giving
r = mv / qB and T = 2πm / qB
Stare at that period for a second, because it is remarkable: T does not contain v. A fast particle sweeps a big circle, a slow one a small circle, and both complete a lap in exactly the same time.
Finally, put an electric field in the picture too. The electric force on a charge is F = qE, and it points along E (for positive q) regardless of whether the charge is moving. The two forces together give the Lorentz force, F = q(E + v × B). Here is how they differ:
| Feature | Electric force qE | Magnetic force qv × B |
|---|---|---|
| Needs the charge to be moving? | No — acts on a charge at rest too | Yes — zero if v = 0 |
| Direction | Along E (positive q) | Perpendicular to both v and B |
| Depends on the angle of travel? | No | Yes — factor sinθ, zero when v ∥ B |
| Work done on the charge | Can be non-zero; changes speed | Always zero; speed unchanged |
| Typical path in a uniform field | Parabola (like projectile motion) | Circle, or a helix if v is slanted |
F = qvB sinθ = (1.6 × 10−19)(2 × 106)(0.5)(sin 30°)
= (1.6 × 10−19)(2 × 106) = 3.2 × 10−13; × 0.5 = 1.6 × 10−13; × 0.5 again
F = 8 × 10−14 N, perpendicular to the plane containing v and B.
Note what happens if the angle were 0°: the force would be exactly zero, no matter how fast the proton travels.
(a) r = mv/qB = (1.67 × 10−27 × 3 × 106) / (1.6 × 10−19 × 0.4)
= (5.01 × 10−21) / (6.4 × 10−20)
r = 7.83 × 10−2 m ≈ 7.8 cm.
(b) T = 2πm/qB = (2π × 1.67 × 10−27) / (6.4 × 10−20) = (1.049 × 10−26)/(6.4 × 10−20)
T = 1.64 × 10−7 s (about 164 ns), i.e. a frequency of 6.10 × 106 Hz.
Double the speed and r doubles to 15.7 cm — but T stays stubbornly at 1.64 × 10−7 s.
Reasoning. Undeflected means the net force is zero, so the electric force exactly cancels the magnetic force: qE = qvB.
The charge q cancels — so the answer is the same for an electron, a proton, or an alpha particle.
v = E/B = 3000 / 0.02
v = 1.5 × 105 m/s.
This is the single most quotable consequence of the Lorentz force: crossed E and B fields pick out one particular speed and let only that speed through unbent.
Force on a Current-Carrying Conductor in a Uniform Magnetic Field
A wire carrying a current is nothing more than an enormous crowd of charges moving in step. If a field pushes each charge sideways, it must push the whole wire sideways. Turning the crowd into a single formula gives
F = I(L × B) → magnitude F = BIL sinθ
Here L is a vector whose length is the length of wire inside the field and whose direction is the direction of the current. And θ is the angle between the wire and the field.
Why it works. Suppose the wire has n free charges per unit volume, each of charge q, drifting with speed vd through a cross-section A. The number of charge carriers in a length L is nAL, and each feels qvdB sinθ. Multiply: F = (nAL)(qvdB sinθ). But the current is exactly I = nAqvd. Substituting collapses the whole thing to F = BIL sinθ. The microscopic picture and the laboratory picture agree, which is always a satisfying moment.
For direction, reach for the same Traffic Cop Grip — only now the thumb points along I instead of v. Nothing else changes. That is the whole point of adopting one gesture.
F = BIL sinθ = 0.6 × 5 × 0.40 × sin 30°
= 0.6 × 5 × 0.40 = 1.2; × 0.5
F = 0.6 N, perpendicular to the plane containing the wire and the field.
Rotate the wire to 90° and the force rises to its maximum, 1.2 N. Lay it along the field and the force vanishes entirely.
Set the two forces equal. BIL = mg
I = mg / BL = (0.020 × 9.8) / (0.8 × 0.50) = 0.196 / 0.40
I = 0.49 A.
Half an ampere lifting a 20 g wire — magnetic forces are not feeble. Do check the current flows the way that makes the palm push upwards; reverse it and the wire is slammed downwards instead.
Force Between Two Parallel Current Carrying Conductors Numericals and the Definition of the Ampere
Put two long parallel wires side by side, a distance d apart, carrying currents I1 and I2. Wire 1 creates a field at the location of wire 2. Wire 2 is a current sitting in that field, so it feels a force. Chain the two results you already have:
- Field of wire 1 at the site of wire 2: B1 = μ0I1 / 2πd.
- Force on a length L of wire 2 sitting in that field, perpendicular to it: F = B1I2L.
- Combine and divide by L to get the force per unit length.
F / L = μ0 I1 I2 / 2πd = 2 × 10−7 × I1I2 / d
The forces on the two wires are equal in size and opposite in direction, exactly as Newton’s third law demands. And the direction is settled by a rule worth naming so you never forget it:
Why it works. Take both currents flowing upwards. Wire 1’s field at wire 2’s position points into the page (right-hand thumb rule). Now apply the Traffic Cop Grip to wire 2: thumb up along I2, fingers into the page along B, and the palm pushes towards wire 1. By the mirror-image argument wire 1 is pushed towards wire 2. They come together. Reverse one current and every step flips once, so the pair fly apart.
F/L = 2 × 10−7 × (10 × 15) / 0.05 = 2 × 10−7 × 150 / 0.05
= 3 × 10−5 / 0.05
F/L = 6 × 10−4 N/m, attractive (same direction → Friends → Attract).
Step 1 — force per metre. F/L = 2 × 10−7 × (20 × 20) / 0.10 = 2 × 10−7 × 4000 = 8 × 10−4 N/m.
Step 2 — multiply by the length. F = 8 × 10−4 × 3
F = 2.4 × 10−3 N, repulsive (opposite directions → Rivals → Repel).
Always read whether the question wants force per metre or total force. Missing the “× 3” is the cheapest mark to lose in the whole chapter.
The definition of the ampere. This formula is not just a numerical exercise — historically it defined the unit of current itself. Put I1 = I2 = 1 A and d = 1 m:
Definition to write in the exam: One ampere is that steady current which, flowing in each of two infinitely long straight parallel conductors of negligible cross-section placed one metre apart in vacuum, produces a force of 2 × 10−7 newton per metre of length on each conductor.
Learn that sentence word for word. It appears as a two-mark question with clockwork regularity, and the marks are given for the specific numbers: 1 m apart, 2 × 10−7 N per metre, in vacuum.
Torque Experienced by a Current Loop in a Uniform Magnetic Field
Here is where the Traffic Cop Grip earns its keep for the third time. Take a rectangular loop of N turns, area A, carrying current I, sitting in a uniform field B with the plane of the loop parallel to the field. Look at the two opposite sides that run perpendicular to B. In one of them the current runs one way; in the other it runs the opposite way. Apply the grip to each: the palm pushes one side up out of the page and the other side down into it.
Two equal forces, opposite in direction, not along the same line — that is precisely the definition of a couple. The net force on the loop is zero, so it does not fly anywhere, but it does twist. Working out the moment of that couple gives
τ = N I A B sinθ
The angle θ here is the angle between B and the normal to the plane of the loop — not between B and the plane itself. Getting this right is worth practising, because it is the single most common slip in the whole topic.
Step 1 — area. A = 0.08 × 0.05 = 4 × 10−3 m2.
Step 2 — the NIAB block. NIAB = 100 × 2 × 4 × 10−3 × 0.5 = 0.4 N·m.
(a) Plane parallel to B means normal perpendicular to B, so θ = 90°, sinθ = 1.
τ = 0.4 N·m (the maximum possible).
(b) θ = 30°, sin 30° = 0.5.
τ = 0.4 × 0.5 = 0.2 N·m.
Useful habit: compute the NIAB block once, then just multiply by whichever sine the question wants. At 60° it would be 0.4 × 0.866 = 0.346 N·m.
Current Loop as a Magnetic Dipole and Its Magnetic Dipole Moment
Look again at the torque formula and group the terms differently: τ = (NIA) × B sinθ. The bracket (NIA) depends only on the loop — how many turns, how much current, how big an area. It says nothing about the field. So let us give it a name. The magnetic dipole moment of a current loop is
m = N I A (unit: A·m2) → τ = m B sinθ, or in vectors τ = m × B
The direction of m is along the normal to the loop, given by curling your right-hand fingers along the current so the thumb points along m. That thumb also points out of the loop’s north face.
Now compare with electrostatics. There, an electric dipole in a field felt a torque τ = pE sinθ. Here, a current loop in a magnetic field feels τ = mB sinθ. Identical structure. That is not a coincidence — a current loop is a magnetic dipole, full stop. A tiny loop and a tiny bar magnet are indistinguishable from a distance; both have a north face and a south face, both line up with an applied field, and both are described by a single vector m.
Step 1. A = 0.08 × 0.05 = 4 × 10−3 m2.
Step 2. m = NIA = 100 × 2 × 4 × 10−3
m = 0.8 A·m2, directed along the normal to the coil.
Step 3. τmax = mB = 0.8 × 0.5
τmax = 0.4 N·m — identical to Example 16(a), as it must be. Two routes, one answer: a reassuring check that you have understood the regrouping.
Moving Coil Galvanometer and Its Current Sensitivity
Everything so far now pays off in a real instrument. A moving coil galvanometer detects and measures small currents, and it is built on exactly one idea: a current loop in a magnetic field twists, and the amount it twists tells you the current.
Construction, in plain words. A rectangular coil of many turns, wound on a light non-magnetic frame, hangs between the poles of a permanent magnet. A phosphor-bronze suspension (or a spiral spring) holds it and provides a restoring twist. A soft iron cylinder sits inside the coil, and the magnet’s pole pieces are carved concave. A light pointer or a mirror-and-lamp arrangement reads the deflection.
Working. Current I through the coil produces a deflecting torque τ = NIAB. The suspension resists with a restoring torque kφ, where φ is the angle turned and k is the torsional constant (torque per unit twist). The needle settles where the two balance:
N I A B = k φ → φ = (NAB / k) I
Look at what that says: the deflection is directly proportional to the current. That is the whole reason the instrument is useful — a linear scale you can read off directly instead of a squashed one you have to interpolate.
Why the radial field matters. The soft iron core and the concave pole pieces together produce a radial magnetic field, meaning the field lines are always along the radius wherever the coil sits. The upshot is that the plane of the coil is always parallel to the field, so sinθ stays equal to 1 no matter how far the coil has turned. Without this, the torque would be NIAB sinθ with a varying θ and the scale would be non-linear and horrible to read. The radial field is not decoration; it is what makes the scale uniform.
Current sensitivity. The deflection produced per unit current is
Current sensitivity = φ / I = N A B / k
To make a galvanometer more sensitive you therefore increase N, A or B, or decrease k by choosing a finer suspension. Phosphor bronze is used precisely because it has a very small torsional constant.
(a) Sensitivity = NAB/k = (50 × 2 × 10−4 × 0.2) / (5 × 10−7)
Numerator = 2 × 10−3. Divide by 5 × 10−7.
Sensitivity = 4000 divisions per ampere.
(b) I = 20 / 4000
I = 5 × 10−3 A = 5 mA.
So this instrument reads comfortably in milliamps — which is exactly why it must be converted before you dare connect it to anything carrying amperes.
Conversion of a Galvanometer into an Ammeter and a Voltmeter
A galvanometer on its own is a delicate thing. It reads only a few milliamps and it has a resistance G of its own. To turn it into a usable meter, you attach exactly one extra resistor — and everything depends on where you attach it.
Making an ammeter. An ammeter must be connected in series and must not disturb the circuit, so its resistance should be as close to zero as possible. Also, the big current I would burn out the coil. Both problems are solved by hanging a small resistance S, called a shunt, in parallel with the galvanometer. The shunt offers an easy bypass, so almost all of the current takes the detour, and only the safe fraction Ig passes through the coil.
Since G and S are in parallel, they share the same potential difference:
IgG = (I − Ig)S → S = IgG / (I − Ig)
Making a voltmeter. A voltmeter is connected across a component and must draw as little current as possible, so its resistance should be enormous. Attach a large resistance R in series with the galvanometer. Now the full-scale current Ig flows only when the total voltage across the combination is V:
V = Ig(R + G) → R = V/Ig − G
| Point of comparison | Ammeter | Voltmeter |
|---|---|---|
| Extra resistor used | Small shunt S | Large resistance R |
| How it is attached to G | In parallel | In series |
| Formula | S = IgG / (I − Ig) | R = V/Ig − G |
| Resistance of the finished meter | GS/(G+S) — very small | R + G — very large |
| Resistance of the ideal version | Zero | Infinite |
| Placed in the circuit | In series with the branch | Across (parallel to) the component |
Step 1. G = 50 Ω, Ig = 2 × 10−3 A, I = 5 A.
Step 2. S = IgG / (I − Ig) = (2 × 10−3 × 50) / (5 − 0.002) = 0.1 / 4.998
S = 0.0200 Ω (about 20 milliohms) connected in parallel.
Step 3 — the ammeter’s own resistance. GS/(G+S) = (50 × 0.020008)/(50.020008)
≈ 0.0200 Ω — essentially the shunt value, which is exactly what we wanted.
Notice how close (I − Ig) is to I. Some questions let you approximate; unless told otherwise, keep the exact subtraction and you cannot go wrong.
Step 1. V/Ig = 10 / (2 × 10−3) = 5000 Ω.
Step 2. R = 5000 − 50
R = 4950 Ω connected in series.
Step 3. Total voltmeter resistance = R + G = 5000 Ω.
Compare the two meters built from the same galvanometer: 0.02 Ω as an ammeter, 5000 Ω as a voltmeter. A factor of a quarter of a million, from one extra resistor placed differently.
Moving Charges and Magnetism Formula List Class 12
Here is the complete moving charges and magnetism formula list class 12 needs — nothing beyond the prescribed scope, nothing missing from it. Read it aloud once on the morning of the paper.
| Situation | Formula | Watch out for |
|---|---|---|
| Biot-Savart law | dB = (μ0/4π) I dl sinθ / r2 | r is element-to-point, not the loop radius |
| Centre of a circular coil | B = μ0NI / 2R | Include N if there are several turns |
| Axis of a circular coil | B = μ0NIR2 / 2(R2+x2)3/2 | Square-root first, then cube |
| Centre of an arc of angle φ | B = μ0Iφ / 4πR | φ must be in radians; semicircle gives μ0I/4R |
| Ampere’s circuital law | ∮ B · dl = μ0Ienclosed | Only currents threading the loop count |
| Long straight wire | B = μ0I / 2πr = 2×10−7 I/r | Falls as 1/r, not 1/r2 |
| Inside a long solenoid | B = μ0nI | n = turns per metre; qualitative only this year |
| Force on a moving charge | F = qvB sinθ; F = q(E + v×B) | Flip the direction for a negative charge |
| Circular path of a charge | r = mv/qB; T = 2πm/qB | T is independent of speed and radius |
| Force on a current-carrying wire | F = BIL sinθ | θ is between the wire and B |
| Two parallel wires | F/L = μ0I1I2 / 2πd | FAR: Friends Attract, Rivals Repel |
| Definition of the ampere | F/L = 2 × 10−7 N/m at d = 1 m | Quote all the conditions for full marks |
| Torque on a current loop | τ = NIAB sinθ = mB sinθ | θ is measured from the normal |
| Magnetic dipole moment | m = NIA (A·m2) | Direction along the normal, right-hand curl |
| Galvanometer deflection | φ = (NAB/k) I | Radial field keeps sinθ = 1 always |
| Current sensitivity | φ/I = NAB/k | Raise N, A or B; lower k |
| Galvanometer → ammeter | S = IgG / (I − Ig) | S must be much smaller than G |
| Galvanometer → voltmeter | R = V/Ig − G | R must be much larger than G |
Moving Charges and Magnetism Class 12 Important Questions
Across recent CBSE papers, the moving charges and magnetism class 12 important questions cluster into six recognisable shapes. If you can handle these six, you can handle almost anything the paper puts in front of you.
- State and apply the Biot-Savart law to find the field at the centre of a circular loop — usually a derivation plus a numerical, three or five marks.
- State Ampere’s circuital law and use it for a straight wire. The derivation is short; write the symmetry argument explicitly, because a mark sits there.
- Define the ampere from the force between two parallel conductors, then compute a force per unit length and name the direction.
- Derive τ = NIAB for a loop, or explain why a current loop behaves as a magnetic dipole with m = NIA.
- Describe the moving coil galvanometer: labelled sketch, working, the role of the radial field, and current sensitivity.
- Convert a galvanometer into an ammeter or a voltmeter of a given range, with a circuit diagram. This is the most frequently repeated numerical in the chapter.
Once you have worked through the sections above, the fastest way to convert understanding into marks is timed practice on a full paper. Work through the CBSE Class 12 Physics sample paper with its marking scheme and watch how magnetism questions are actually phrased and awarded — you will notice how many marks go to a single well-drawn diagram or one correctly stated direction. If diagrams are where you lose marks, the habits in the ray-diagram discipline from the Class 10 Light chapter transfer directly: label everything, mark directions with arrows, and never leave a sketch unannotated.
What’s NOT in Your 2026-27 Syllabus
- The cyclotron — construction, working, resonance condition, maximum energy. Not in scope.
- The toroid — Ampere’s law applied to a toroidal coil. Not in scope.
- The velocity selector as a named topic — you should still be able to balance electric and magnetic forces on a moving charge (that is in scope, and it is Example 10 above), but the device does not need to be studied or named.
- Voltage sensitivity of a galvanometer — only current sensitivity is prescribed.
- Deriving B = μ0nI for a solenoid using Ampere’s law — the solenoid is explicitly qualitative treatment only. Know what it is, sketch its field, quote the result; skip the rectangular Amperian loop derivation.
Trimming these frees real time for the topics that are examined — especially the galvanometer conversions.
Practice Worksheet
Ten original questions, in roughly increasing difficulty. Attempt each one on paper before you open the answer. Take μ0 = 4π × 10−7 T·m/A, e = 1.6 × 10−19 C, and mass of electron = 9.1 × 10−31 kg.
Q1. A circular coil of 20 turns and radius 5 cm carries a current of 1.5 A. Find the magnetic field at its centre.
Show Answer
Numerator = 4π × 10−7 × 30 = 3.770 × 10−5. Divide by 0.1.
B = 3.77 × 10−4 T, perpendicular to the plane of the coil.
Q2. A long straight wire carries 12 A. Calculate the magnetic field at a perpendicular distance of 3 cm from it.
Show Answer
B = 8 × 10−5 T, tangential to the circle of radius 3 cm around the wire.
Q3. An electron moves at 5 × 106 m/s at right angles to a uniform magnetic field of 0.3 T. Find the magnitude of the force on it, and state whether its speed changes.
Show Answer
F = 2.4 × 10−13 N.
The speed does not change. The force is always perpendicular to the velocity, so it does zero work and only changes the direction of motion.
Q4. An electron moving at 2 × 106 m/s enters a uniform magnetic field of 0.05 T at right angles. Find the radius of its circular path.
Show Answer
= (1.82 × 10−24) / (8 × 10−21)
r = 2.275 × 10−4 m ≈ 0.23 mm.
Q5. A straight wire 25 cm long carries 6 A and lies perpendicular to a uniform field of 0.4 T. Find the force on it. What would the force become if the wire were turned to lie along the field?
Show Answer
F = 0.6 N.
Turned along the field, θ = 0°, sinθ = 0, so F = 0.
Q6. Two long parallel wires 4 cm apart carry 8 A and 12 A in opposite directions. Find (a) the force per unit length between them and (b) the total force on a 2 m length. State whether it is attractive or repulsive.
Show Answer
F/L = 4.8 × 10−4 N/m.
(b) F = 4.8 × 10−4 × 2 = 9.6 × 10−4 N.
Opposite directions means Rivals, so the force is repulsive.
Q7. A rectangular coil of 200 turns measuring 4 cm × 3 cm carries 0.5 A in a uniform field of 0.8 T. Find (a) its magnetic dipole moment, (b) the maximum torque on it, and (c) the torque when the normal to the coil makes 30° with the field.
Show Answer
(a) m = NIA = 200 × 0.5 × 1.2 × 10−3 = 0.12 A·m2.
(b) τmax = mB = 0.12 × 0.8 = 0.096 N·m.
(c) τ = mB sin 30° = 0.096 × 0.5 = 0.048 N·m.
Q8. A galvanometer of resistance 100 Ω shows full-scale deflection for 1 mA. What shunt is needed to convert it into an ammeter of range 0–2 A?
Show Answer
S = 5.00 × 10−2 Ω ≈ 0.05 Ω, connected in parallel with the galvanometer.
Size check: S is far smaller than G — correct for a shunt.
Q9. A galvanometer of resistance 60 Ω gives full-scale deflection for 5 mA. Find the resistance to be connected, and how, to convert it into a voltmeter of range 0–15 V. What is the total resistance of the voltmeter?
Show Answer
R = 3000 − 60 = 2940 Ω, connected in series.
Total voltmeter resistance = R + G = 3000 Ω.
Q10. A galvanometer coil has 30 turns, area 1.5 × 10−4 m2, and sits in a radial field of 0.25 T. Its suspension has torsional constant 1.5 × 10−7 N·m per division. Find its current sensitivity, and explain why the field is made radial.
Show Answer
Numerator = 1.125 × 10−3. Divide by 1.5 × 10−7.
Current sensitivity = 7500 divisions per ampere.
Why radial? A radial field keeps the plane of the coil always parallel to the field, so sinθ stays equal to 1 through the whole swing. The deflecting torque is then NIAB regardless of position, the deflection stays directly proportional to the current, and the scale is uniform and easy to read.
You have just met the first genuinely three-dimensional idea in Class 12 Physics, and you handled it. Look back at the Traffic Cop Grip: it carried you through the force on a charge, the force on a wire, the pull between two wires, and the twist on a loop — four topics, one gesture. That is what real understanding looks like, and it is far more durable than four separate memorised rules.
So do not aim for perfect today. Aim for one more correct answer than yesterday. Ten questions today, eleven tomorrow, twelve the day after — and by the week’s end this chapter will have quietly become one of your strong ones.