Which statements about electromagnetic quantities are correct?

D = εE in a linear isotropic medium

E = −∇V in an electrostatic field

B is flux density, while magnetic flux is its surface integral

D = εE in a linear isotropic medium

B and H are always numerically equal

E = −∇V in an electrostatic field

B is flux density, while magnetic flux is its surface integral

The Poynting vector represents:

Electromagnetic power flow per unit area

Electricity and Magnetism: Fields, Induction and Magnetic Circuits

Section 2 of the GATE Instrumentation Engineering (IN) paper. Instrumentation students meet electromagnetism because sensors, actuators, inductive elements and measurement circuits all sit on the same physical foundation. Start with the field picture: charge creates an electric field, current creates a magnetic field, and changing magnetic flux creates an emf. Once that chain is clear, the equations become tools rather than isolated formulas.

Electric field, flux and Gauss law

For a point charge, electric field falls as 1/r². Gauss law is most useful when symmetry lets you turn flux into field magnitude: spherical, cylindrical or planar symmetry. Do not use it merely because the word "charge" appears; without symmetry the surface integral may be no easier.

Quantity	Meaning	Key relation
E	Field intensity	F = qE
D	Flux density	D = εE
V	Potential	E = −∇V

Magnetic field and magnetic circuits

Biot–Savart and Ampere answer different versions of the same practical question: what magnetic field does current create? Ampere is the shortcut only when symmetry makes ∮H·dl easy. In a magnetic circuit, mmf NI drives flux through reluctance, much like voltage drives current through resistance. The analogy is helpful, but remember that leakage flux and nonlinear B-H behaviour make the magnetic circuit imperfect.

⚠️ Exam trap

B, H and flux are not interchangeable. H is linked to current distribution; B additionally depends on permeability. Always check the medium before replacing one by another.

Faraday, Lorentz force and Maxwell equations

Faraday law is the bridge from magnetism to instrumentation: changing flux creates induced emf, while Lorentz force explains the force on a moving charge. Maxwell’s equations unify the static and dynamic cases. For exam problems, first identify whether the source is charge, current or time-varying flux; that choice usually tells you which equation to start with.

Law	Differential idea	Use
Gauss electric	Divergence of D follows charge density	Charge-field relation
Faraday	Curl E follows changing B	Induced emf
Ampere-Maxwell	Curl H follows current and changing D	Dynamic magnetic field

Boundary conditions and electromagnetic energy

At an interface, normal flux density is controlled by free surface charge, while tangential magnetic field is controlled by surface current. These conditions are often faster to use than re-solving the full field equations. Electromagnetic power flow is represented by the Poynting vector S = E × H; its direction tells you where energy is flowing.

🧠 Memory hook

Normal D notices free charge; tangential H notices surface current.

Key takeaways

Choose Gauss or Ampere from symmetry, not from keywords.
Keep E, D, B, H and flux conceptually separate; constitutive relations connect them.
A changing magnetic flux can induce emf; this is central to inductive sensing and transformers.
Use boundary conditions and the Poynting vector when the question is about interfaces or energy flow.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

For a point charge in a homogeneous medium, the magnitude of electric field varies with distance r as:
1. r
2. 1/r
3. 1/r²
4. r²
Show answer
Answer: C — 1/r²
Coulomb’s law gives E proportional to 1/r².
Which law is generally the most convenient for finding the electric field of a uniformly charged infinite plane?
1. Gauss law
2. Faraday law
3. Lenz law
4. Biot–Savart law
Show answer
Answer: A — Gauss law
Planar symmetry makes a Gaussian pillbox reduce the flux calculation to a simple algebraic relation.
A magnetic circuit has N = 200 turns carrying 0.5 A. Its magnetomotive force is:
1. 50 AT
2. 100 AT
3. 200 AT
4. 400 AT
Show answer
Answer: B — 100 AT
MMF = NI = 200 × 0.5 = 100 ampere-turns.
Which statements about electromagnetic quantities are correct?
1. D = εE in a linear isotropic medium
2. B and H are always numerically equal
3. E = −∇V in an electrostatic field
4. B is flux density, while magnetic flux is its surface integral
Show answer
Answer: A — D = εE in a linear isotropic medium; C — E = −∇V in an electrostatic field; D — B is flux density, while magnetic flux is its surface integral
The constitutive relation depends on the medium; B and H are not universally equal. Electric potential gives E through its negative gradient in electrostatics, and flux is the surface integral of B.
If magnetic flux through a single-turn loop changes from 0.02 Wb to 0.01 Wb in 0.005 s, the magnitude of average induced emf is:
1. 0.5 V
2. 1 V
3. 2 V
4. 4 V
Show answer
Answer: C — 2 V
Faraday's law gives |e| = |ΔΦ|/Δt. The flux changed by 0.01 − 0.02 = −0.01 Wb, so |e| = 0.01/0.005 = 2 V. The 4 V option comes from using the initial flux 0.02 Wb in place of the change in flux, which is the usual slip: an induced emf responds to how much the flux moved, never to how much of it there was.
The Poynting vector represents:
1. Electric charge density
2. Electromagnetic power flow per unit area
3. Magnetic reluctance
4. Stored electric charge
Show answer
Answer: B — Electromagnetic power flow per unit area
S = E × H gives electromagnetic power flow density and its direction.
For a boundary with no free surface charge, which normal-component condition follows from Gauss law?
1. Dn1 = Dn2
2. Et1 = −Et2
3. Bn1 = 0 always
4. Ht1 = Ht2 always
Show answer
Answer: A — Dn1 = Dn2
The normal component of D changes by the free surface charge density, so with zero free surface charge it is continuous.
An air-core solenoid has 1000 turns and carries 2 A. If its ideal magnetic-circuit length is 0.5 m, what is NI/l in A/m?
Numerical answer — type the value.
Show answer
Answer: 4000
For an ideal magnetic circuit the magnetic field intensity is H = NI/l, so H = (1000 × 2)/0.5 = 4000 A/m. The quantity in the denominator is the mean magnetic path length in metres, not the number of turns — dividing by N instead is the common error and yields 2 A/m.