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Magnetic Fields: Question 5

Syllabus 20.5

Structured A2 8 marks

A flat, rectangular search coil has N=250N=250 turns and a cross-sectional area of A=4.0×10−4 m2A=4.0\times10^{-4}\text{ m}^2. The coil is initially held with its plane perpendicular to a uniform magnetic field of flux density B1=0.080 TB_1=0.080\text{ T}, so that the field passes straight through the coil.

(a) State what is meant by magnetic flux, and calculate the magnetic flux Φ\Phi through the coil in this initial position. [2]

(b) Calculate the flux linkage of the coil in this initial position. [1]

(c) The coil is then pulled out of the field over a time interval of 0.050 s0.050\text{ s}, during which the flux density through the coil falls uniformly from 0.080 T0.080\text{ T} to zero. Use Faraday's law to calculate the average e.m.f. induced in the coil during this time. [3]

(d) State Lenz's law, and use it to describe the direction of the induced current in the coil, relative to the original magnetic field, while the coil is being removed. [2]

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Worked solution

Part (a): Magnetic flux

Magnetic flux through a surface is defined as the product of the magnetic flux density normal to the surface and the area of the surface. Since the coil’s plane is perpendicular to the field (so the field is entirely normal to it), Φ=B1A\Phi=B_1A: Φ=0.080×4.0×10−4=3.2×10−5 Wb\Phi=0.080\times4.0\times10^{-4}=3.2\times10^{-5}\text{ Wb}

Recomputing as a check: 8.0×4.0=328.0\times4.0=32, and the powers of ten give 10−2−4=10−610^{-2-4}=10^{-6}, so Φ=32×10−6=3.2×10−5 Wb\Phi=32\times10^{-6}=3.2\times10^{-5}\text{ Wb}, the same result.

Part (b): Flux linkage

Flux linkage is the total flux linked with all the turns of the coil, NΦN\Phi: NΦ=250×3.2×10−5=8.0×10−3 WbN\Phi=250\times3.2\times10^{-5}=8.0\times10^{-3}\text{ Wb}

Recomputing as a check: 250×3.2=800250\times3.2=800, and 800×10−5=8.0×10−3 Wb800\times10^{-5}=8.0\times10^{-3}\text{ Wb}, consistent.

Part (c): Average e.m.f. from Faraday’s law

Faraday’s law states that the induced e.m.f. equals the rate of change of flux linkage: ε=∣Δ(NΦ)Δt∣\varepsilon=\left|\frac{\Delta(N\Phi)}{\Delta t}\right|

The flux linkage falls uniformly from 8.0×10−3 Wb8.0\times10^{-3}\text{ Wb} (part (b)) to zero (since BB falls to zero) over Δt=0.050 s\Delta t=0.050\text{ s}: ε=8.0×10−3−00.050=0.16 V\varepsilon=\frac{8.0\times10^{-3}-0}{0.050}=0.16\text{ V}

Recomputing as a check: 8.0×10−3÷5.0×10−2=(8.0÷5.0)×10−3−(−2)=1.6×10−1=0.16 V8.0\times10^{-3}\div5.0\times10^{-2}=(8.0\div5.0)\times10^{-3-(-2)}=1.6\times10^{-1}=0.16\text{ V}, the same result.

Part (d): Lenz’s law and the direction of the induced current

Lenz’s law states that the direction of an induced e.m.f. (and any resulting current) is always such as to oppose the change in flux that produces it.

Here, the flux through the coil is decreasing as it is pulled out of the field. By Lenz’s law, the induced current must flow in the direction that opposes this decrease. That is, the current flows so that, by the right-hand grip rule, its own magnetic field inside the coil points in the same direction as the original external field, trying to maintain the flux that is being lost. This is consistent with energy conservation: the induced current opposes the motion removing the coil (via a retarding force on it), so work must be done against this force to pull the coil out, and that work is the source of the electrical energy dissipated by the induced current.

Final answers

  • (a) Magnetic flux is flux density ×\times area normal to the field; Φ=3.2×10−5 Wb\Phi=\boxed{3.2\times10^{-5}}\text{ Wb}
  • (b) Flux linkage NΦ=8.0×10−3 WbN\Phi=\boxed{8.0\times10^{-3}}\text{ Wb}
  • (c) Average e.m.f. ε=0.16 V\varepsilon=\boxed{0.16}\text{ V}
  • (d) By Lenz’s law, the induced current flows so as to oppose the decreasing flux, creating its own field inside the coil in the same direction as the original field