3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
In vacuum, the electric field of a plane electromagnetic wave is E=(30i^−40j^)cos(ωt+kz)V/m, where k=1.0×107rad/m and ω=ck, with c=3.0×108m/s; i^,j^,k^ are unit vectors along x,y,z. The magnetic field of the wave, in tesla, is ______. Exactly one option is correct.
A(1.33i^+1.00j^)×10−7cos(ωt+kz)
B(−1.33i^−1.00j^)×10−7cos(ωt+kz)
C(−1.00i^−1.33j^)×10−7cos(ωt+kz)
D(−1.33i^−1.00j^)×10−7cos(ωt−kz)
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Answer: B
The phase ωt+kz stays constant when z decreases as t increases, so the wave travels along −k^; the phase of B is the same, cos(ωt+kz) (this rules out (D)).
For a plane wave in vacuum, B=c1n^×E with n^=−k^ the propagation direction.
−k^×(30i^−40j^)=−30(k^×i^)+40(k^×j^)=−30j^−40i^.
So B=3.0×108−40i^−30j^cos(ωt+kz)=(−1.33i^−1.00j^)×10−7cos(ωt+kz)T; magnitude 50/c=1.67×10−7T.
Check: E⋅B=30(−40)+(−40)(−30)=0 and E×B∝(30)(−30)−(−40)(−40)=−2500 along k^, i.e. along −k^, as required.
Trap: (A) is the field for a wave moving along +z; (C) swaps the components and is not even perpendicular to E.
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A parallel-plate capacitor in vacuum has circular plates of radius R; the plate separation is much smaller than R and fringing is negligible. It is being charged by a constant current I carried by long straight wires along the common axis of the plates. Consider points in the mid-plane of the gap at perpendicular distance r from the axis, both inside (r<R) and outside (r>R) the plate edge. Let Bmax be the largest magnitude of the magnetic field at such points. The set of all such points where the magnetic field has magnitude Bmax/2 is ______. Exactly one option is correct.
Ar=R/2 and r=2R only
Br=R/2 only
Cr=R/2 and r=2R only
Dr=R/4 and r=4R only
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Answer: A
Use the Ampere-Maxwell law ∮B⋅dl=μ0(Ic+ε0dtdΦE) on a circle of radius r in the mid-plane, centred on the axis; by symmetry B is tangential and uniform on it. Take the flat disc bounded by the circle as the surface: no conduction current crosses it.
The field between the plates is uniform, E=q/(ε0πR2), so the total displacement current is ε0πR2dE/dt=dq/dt=I, spread uniformly over the plate area.
Inside (r<R): B⋅2πr=μ0Ir2/R2, so B=2πR2μ0Ir (rises linearly).
Outside (r>R): the disc collects all the displacement current: B=2πrμ0I (falls as 1/r).
Maximum at r=R: Bmax=2πRμ0I. Half of it: inside r/R=1/2, outside R/r=1/2. So r=R/2 and r=2R.
Insight: the gap is not a "hole" in the circuit: the displacement current makes the field outside the gap identical to that of an unbroken wire, and inside it grows like the field inside a thick wire.
Difficulty: Very hard
Question 3Level 3 · AdvancedSingle correct
A plane electromagnetic wave travels through a transparent, homogeneous, non-magnetic (μ=μ0) dielectric. Its electric field is E=60i^sin(kz−ωt)V/m with k=2.0×107rad/m and ω=4.5×1015rad/s. Take c=3.0×108m/s. Take the vacuum-wavelength ranges as: ultraviolet 10 nm to 400 nm, visible 400 nm to 700 nm, infrared 700 nm to 1 mm. The relative permittivity of the dielectric, the amplitude of the magnetic field of the wave in the dielectric, and the band to which this radiation belongs (judged by its wavelength in vacuum) are, respectively, ______. Exactly one option is correct.
A4/3; 2.0×10−7T; visible
B16/9; 2.0×10−7T; ultraviolet
C16/9; 2.67×10−7T; ultraviolet
D16/9; 2.67×10−7T; visible
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Answer: D
Wave speed in the medium: v=ω/k=4.5×1015/2.0×107=2.25×108m/s, so n=c/v=4/3.
Non-magnetic medium: v=1/μ0εrε0=c/εr, so εr=n2=16/9.
In the medium E0=vB0: B0=60/2.25×108=2.67×10−7T (not E0/c).
Wavelength in the medium λm=2π/k=314nm; in vacuum (same frequency) λ0=nλm=419nm: visible (violet).
Trap: classifying by the in-medium wavelength 314 nm gives "ultraviolet"; using E0/c gives 2.0×10−7 T; taking εr=n gives 4/3. The band is a property of the frequency, which does not change in the medium.
Difficulty: Very hard
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