01Galvanometer
Converting a galvanometer into an ammeter
A shunt in parallel diverts the excess current; both branches share one voltage.
IgG = (I − Ig)S
S = IgGI − Ig = 0.01 × 1000.99
D S = 1.01 Ω
Worked questions
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02Solenoid
Second solenoid, half the turns and a third of the current
Ratio method — units of n cancel, so turns/cm need no conversion.
B₂B₁ = n₂n₁·i₂i₁ = 12·13 = 16
B 1.05 × 10−2 Wb/m2
03Magnetic moment
Magnetic moment of a spinning charged disc
Charge sits on an area, so split into rings and integrate.
dI = σω x dx , dM = dI(πx2)
M = σωπR44 = qωR24
A M = ¼qωR2
04Force on a loop
Rectangular loop beside a long wire
Near side attracted in a strong field beats far side repelled in a weak one.
F = μ₀IiL2π(1d₁ − 1d₂) > 0
C Moves toward the wire
05Torque & work
Work to rotate a coil through 180°
Equilibrium means θ₁ = 0, so the turn costs the full 2MB.
M = NiA = 50 × 2 × π(0.04)2
W = MB(cos θ₁ − cos θ₂) = 2MB
A W ≈ 0.1 J
06Constants
Field written through ε₀ and c
Swap μ₀ for 1/ε₀c². The 9 from the current cancels the 9 from c².
B = μ₀I2R = 9μ₀ , μ₀ = 1ε₀c2
A B = 1 / (ε₀1016)
07Two coils
Concentric coils, currents in opposite senses
Coplanar and opposing, so subtract; direction follows the stronger coil.
BA = μ₀(24)(10)2(0.25) = 480μ₀
BB = μ₀(18)(15)2(0.15) = 900μ₀
D B = 420μ₀ T
08Direction
Field below a power line
Right-hand grip: thumb east, fingers curl — below the wire they point north.
direction of B ∝ Î × r̂ = east × down = north
B South to North
09Same wire
Same 20 cm wire wound as 4 turns vs 2 turns
Length fixed ⇒ R = L/2πN ⇒ B ∝ N². The length never enters the ratio.
B = πμ₀N2IL
⇒
BABB = (42)2
A 4 : 1
10Symmetry
Symmetry of the field around a straight wire
B depends only on perpendicular distance — that is what lets B leave the integral.
∮B⃗·dl⃗ = B(2πr) = μ₀I
B Cylindrical symmetry
11Ampère’s law
Inside a thin-walled hollow tube
All the current is on the wall, so a loop inside encloses nothing.
Ienc = 0 ⇒ B × 2πr = 0
(solid wire would give μ₀Ir2πa2)
C Zero
12Assertion & Reason
Radial field in a moving-coil galvanometer
Concave poles keep θ = 90° always, so the scale reads current linearly.
τ = NIAB sin θ , sin 90° = 1
NIAB = kφ ⇒ I ∝ φ
A Both true, R explains A
13Time period
What is the time period independent of?
m, q and B all appear. v and r do not.
T = 2πmqB
A Speed
14Cross product
Charge with v along ĵ in a field along î
Order matters: ĵ × î = −k̂, the reverse of î × ĵ = +k̂.
F⃗ = q(vĵ) × (Bî) = −qvB k̂
A Z-axis (force along −Z)
15Axial field
Radius from a 216 : 125 field ratio
216 = 6³ and 125 = 5³ — take the cube root first, then solve.
(R2+27R2+16)3/2 = 216125
25R2+675 = 36R2+576
A R = 3 cm
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23Circular path
Area enclosed by the path
A ∝ E and A ∝ 1/q². It goes as v², not v — and as T², not 1/T.
A = πr2 = πm2v2q2B2 = 2πmEq2B2
B Statements i & ii only
24Tangent galvanometer
Deflection of the needle
Coil field ⟂ Earth’s horizontal field; the needle sits along the resultant.
tan θ = BcoilBH = μ₀NI2RBH
= 4×10−54×10−5 = 1
B θ = 45°
25Non-conservative
Pole carried once around a wire
Closed path, yet the work isn’t zero — the field of a current has no potential.
W = m∮B⃗·dl⃗ = μ₀mI
= 8 × 4π×10−7 × 5
B W = 16π × 10−6 J
Formula bank for this set
New material beyond Set 1Galvanometer conversions
- To an ammeter — shunt in parallel
- S = IgGI − Ig (make it small)
- To a voltmeter — resistance in series
- R = VIg − G (make it large)
- Current split between the branches
- IgI = SS + G
- Resistance of the finished ammeter
- GSG + S
- Tangent galvanometer
- tan θ = μ₀NI2RBH ⇒ I = k tan θ
Magnetic moment, torque, work
- Moment of a current loop
- M = NIA
- Rotating charged ring / disc
- Mring = ½qωR2 Mdisc = ¼qωR2
- Torque in a field
- τ = MB sin θ (zero at 0° and 180°)
- Potential energy
- U = −MB cos θ
- Work to rotate
- W = MB(cos θ₁ − cos θ₂)
0→90°: MB · 0→180°: 2MB · 90→180°: MB - Moving-coil galvanometer
- NIAB = kφ ⇒ I ∝ φ (radial field keeps sin θ = 1)
Where the field is zero
- Inside a thin-walled hollow tube
- B = 0 (no enclosed current)
- Open hole of a toroid, outside a solenoid
- B = 0
- On the line of a straight segment
- dl⃗ ∥ r̂ ⇒ contribution 0
- Radial arms of a loop about its centre
- B = 0
- Stationary charge in a magnetic field
- F = 0 (v = 0)
Vectors & constants
- Cyclic cross products
- î × ĵ = k̂ · ĵ × k̂ = î · k̂ × î = ĵ
reverse the order and the sign flips - Compass cross products (positive charge)
- north × down = west · north × up = east
east × down = north · north × east = up - The magnetism–electrostatics bridge
- μ₀ε₀c2 = 1 ⇒ μ₀ = 1/ε₀c2
- Work on a pole encircling a current
- W = μ₀mI per lap (non-conservative)
- Area of the circular path
- A = 2πmEq2B2
Ideas that carried this set
What to recognise on sight- Ratios beat arithmetic. Q2, Q7, Q9 and Q15 all collapse when you divide the two cases: μ₀, I and even unit conversions cancel before you touch a calculator.
- “Same wire” means B ∝ N². Q9. Fixed length forces R = L/2πN, so the turns count enters twice. Watch for “same material and same length”.
- Spread-out charge needs integration. Q3. A ring is one loop; a disc is infinitely many, and the outer rings carry more area — which is why the disc gives half the ring’s moment.
- Instrument design is physics with a purpose. Q1, Q12, Q24. The shunt exists to protect the coil, the radial field exists to make the scale linear, and the tangent law exists to turn an angle into a current reading.
- Magnetism is not conservative. Q25. ∮B·dl = μ₀I ≠ 0, so carrying a pole in a closed loop around a wire does real work every lap. “Closed path ⇒ zero” is an electrostatics habit that fails here.
Traps in this set
Where the marks leak| The mistake | In | Fix |
|---|---|---|
| Putting the shunt in series | Q1 | Ammeter = parallel shunt; voltmeter = series resistance. |
| Treating a disc like a single ring | Q3 | Integrate: disc gives ¼qωR², not ½. |
| Using W = MB instead of 2MB | Q5 | Equilibrium → 180° spans cos 0 to cos 180. |
| Adding the two coil fields | Q7 | Opposite senses subtract: 900 − 480, not 1380. |
| Reading the field above the wire instead of below | Q8 | The field reverses across the wire. |
| Applying the solid-wire formula to a hollow tube | Q11 | “Thin walled” ⇒ nothing enclosed ⇒ zero. |
| Getting the cross-product order backwards | Q14 | ĵ × î = −k̂, not +k̂. |
| Answering “zero” for a closed path | Q25 | Magnetic fields of currents are non-conservative. |