University of Burdwan · CBCS · PYQs 2018–2024

PHYS7021 · Electricity, Magnetism & Wave Optics

Sem-VII Physics Minor — syllabus-wise theory ➜ solved past questions ➜ model answers · Indian-textbook style · mobile friendly

Section 2 — Magnetostatics

GE-2 / PHYS7021 unit 2 · 4 lecture-hours · 6 solved questions

2.1 Theory you need

A. Biot–Savart law and applications

dB = (μ0/4π) · I dl × r̂ / r2 ,   μ0 = 4π×10−7 H/m
I (up) B (anticlockwise seen from top) r from wire

B. Divergence & curl of B; Ampère's law; vector potential

∇·B = 0 (no magnetic monopoles) · ∮B·dl = μ0Ienc → ∇×B = μ0J · B = ∇×A
loop radius R, current I axis, distance x P (B along axis)

2.2 Solved past questions — exam-style answers

1. What is magnetic vector potential? (2018 adds: given A = r2î, calculate B.)
×220182022
  1. Definition: A is the vector field whose curl gives the magnetic field: B = ∇×A. It exists because ∇·B = 0; unit Wb/m (or T·m).
  2. Numerical: A = r2î with r2 = x2+y2+z2; only Ax ≠ 0.
  3. Bx = ∂Az/∂y − ∂Ay/∂z = 0.
  4. By = ∂Ax/∂z − ∂Az/∂x = 2z − 0 = 2z.
  5. Bz = ∂Ay/∂x − ∂Ax/∂y = 0 − 2y = −2y.

∴ B = 2z ĵ − 2y k̂

2. State Ampère's circuital law; write its mathematical form / show ∇×B = μ0J.
×3201820192023
  1. Statement: the line integral of B around any closed loop equals μ0 times the net current threading the loop: ∮B·dl = μ0Ienc.
  2. Write Ienc = ∫J·dA and use Stokes' theorem: ∮B·dl = ∫(∇×B)·dA.
  3. Then ∫(∇×B − μ0J)·dA = 0 for every surface ⇒ ∇×B = μ0J (differential form).

∴ ∮B·dl = μ₀I ; ∇×B = μ₀J

3. State Biot–Savart's law in magnetostatics.
×3201820192022
  1. Statement: the magnetic field due to a current element I dl at distance r is dB = (μ0/4π) · I dl × r̂ / r2.
  2. Magnitude: dB = (μ0/4π)(I dl sinθ)/r2 — directly proportional to current, element length and sinθ; inversely to r2.
  3. Direction: perpendicular to the plane of dl and r̂ (right-hand rule). It is the magnetic analogue of Coulomb's law.

∴ dB = (μ₀/4π) I dl×r̂ / r²

4. Derive B at distance r from a long straight conductor (2023 numerical: I = 1.5 A, r = 3 cm).
×3201920222023
  1. By symmetry B is constant in magnitude along a circle of radius r around the wire and tangent to it.
  2. Ampère's law: ∮B dl = B(2πr) = μ0I ⇒ B = μ0I/2πr.
  3. Numerical: B = (2×10−7 × 1.5)/0.03 = 3×10−7/3×10−2.
  4. = 1×10−5 T (= 0.1 gauss).

∴ B = μ₀I/2πr ; numerical: 1×10⁻⁵ T

5. Applying Biot–Savart's law derive B at a point on the axis of a current-carrying circular loop.
×220182023
  1. For every element, dl ⊥ r̂, so dB = (μ0/4π) I dl/(R2+x2).
  2. Resolve dB: the components perpendicular to the axis cancel in pairs; axial components add with factor sinφ = R/√(R2+x2).
  3. B = ∫dB sinφ = (μ0/4π)·(I R/(R2+x2)3/2)·∮dl, with ∮dl = 2πR.
  4. B = μ0I R2/2(R2+x2)3/2; at centre (x=0): B = μ0I/2R.

∴ B(axis) = μ₀IR²/2(R²+x²)^{3/2} ; centre: μ₀I/2R

6. What is the meaning of ∇·B = 0? Using Biot–Savart law prove ∇·B = 0.
2022
  1. Meaning: magnetic field lines are closed loops; no isolated magnetic poles (monopoles) exist; net magnetic flux through any closed surface is zero.
  2. Proof: from Biot–Savart, B(r) = (μ0I/4π)∮ dl′ × (r−r′)/|r−r′|3.
  3. Since (r−r′)/|r−r′|3 = −∇(1/|r−r′|), the field can be written as B = ∇×A with A = (μ0I/4π)∮ dl′/|r−r′|.
  4. Divergence of a curl is identically zero ⇒ ∇·B = 0.

∴ ∇·B = 0 — no magnetic monopoles

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