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Questions on magnetic flux and flux linkage
Magnetic flux is the field passing through an area, Φ = BA in webers, and flux linkage NΦ counts that flux once for every turn of the coil. Tilt the coil and only the field component along its normal counts, giving NΦ = BAN cos θ, with θ measured to the normal. A search coil and an oscilloscope can confirm the cos θ dependence.
18 original questions · 49 marks · the magnetic flux and flux linkage notes · Magnetic fields
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State the equation for the magnetic flux through an area whose normal is parallel to the field, and give the unit of magnetic flux.
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Φ = BA, where B is the flux density and A the area perpendicular to the field (1); unit: weber (Wb) (1).A magnetic field of flux density 0.20 T passes normally through an area of 0.05 m2. Calculate the magnetic flux through the area.
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Φ = BA = 0.20 × 0.05 (1)
Φ = 0.01 Wb (1)Define the flux linkage of a coil and state its unit.
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The product of the magnetic flux through the coil and the number of turns, NΦ (1); unit: weber-turns (1).A circular loop of wire of diameter 8.0 cm lies with its plane perpendicular to a magnetic field of flux density 0.15 T. Calculate the magnetic flux through the loop.
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A = πr2 = π × 0.0402 = 5.0 × 10−3 m2 (1)
Φ = BA = 0.15 × 5.0 × 10−3 = 7.5 × 10−4 Wb (1)State the angle between the normal to a coil and a magnetic field at which the flux linkage of the coil is half its maximum value.
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Flux linkage is proportional to cosθ, and cosθ = 0.5 at θ = 60° (1).The number of turns of a coil is doubled, the flux density it sits in is doubled and its area is halved. State and explain the effect on the flux linkage of the coil.
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Flux linkage NΦ = BAN cosθ is proportional to B, A and N (1); the factor is 2 × ½ × 2 = 2, so the flux linkage doubles (1).A coil of 200 turns and area 0.02 m2 is placed with its plane perpendicular to a magnetic field of flux density 0.15 T. Calculate the flux linkage of the coil.
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NΦ = BAN = 0.15 × 0.02 × 200 (1)
NΦ = 0.60 Wb-turns (1)A coil of 100 turns and area 0.01 m2 is placed in a magnetic field of flux density 0.30 T with the normal to its plane at 60° to the field. Calculate the flux linkage of the coil.
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NΦ = BAN cosθ = 0.30 × 0.01 × 100 × cos 60° (1)
NΦ = 0.15 Wb-turns (1)A magnetic flux of 8.0 × 10−3 Wb passes normally through an area of 0.04 m2. Calculate the magnetic flux density.
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B = Φ/A = (8.0 × 10−3)/0.04 (1)
B = 0.20 T (1)A coil is rotated in a uniform magnetic field. Describe how the flux linkage varies with the orientation of the coil, and state the orientations that give maximum and zero flux linkage.
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NΦ = BAN cosθ, where θ is the angle between the normal to the coil and the field (1). Maximum when θ = 0, with the plane of the coil perpendicular to the field (1); zero when θ = 90°, with the plane of the coil parallel to the field (1).A 400-turn coil of area 1.2 × 10−3 m2 is placed in a magnetic field of flux density 0.25 T. The flux linkage of the coil is 0.060 Wb-turns. Calculate the angle between the normal to the coil and the field.
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cosθ = NΦ/(BAN) (1)
= 0.060/(0.25 × 1.2 × 10−3 × 400) = 0.060/0.12 = 0.50 (1)
θ = 60° (1)At a certain location the Earth's magnetic field has flux density 49 μT and points at 66° below the horizontal. A 2000-turn coil of area 0.030 m2 lies flat on a horizontal bench. Calculate the flux linkage of the coil.
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The normal to the coil is vertical, so the angle between the field and the normal is 90° − 66° = 24° (1)
NΦ = BAN cosθ = 49 × 10−6 × 0.030 × 2000 × cos 24° (1)
NΦ = 2.7 × 10−3 Wb-turns (1)A 250-turn coil of area 6.0 × 10−4 m2 sits in a magnetic field of flux density 0.32 T with its plane at 35° to the field. Calculate the flux linkage of the coil. Go on to calculate the change in flux linkage when the coil is rotated so that it is face-on to the field.
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The normal to the coil is at 90° − 35° = 55° to the field (1)
NΦ = BAN cos 55° = 0.32 × 6.0 × 10−4 × 250 × cos 55° = 0.028 Wb-turns (1)
Face-on: NΦ = BAN = 0.048 Wb-turns (1)
Change = 0.048 − 0.028 = 0.020 Wb-turns (1)A coil of 50 turns and area 0.02 m2 is placed in a magnetic field of flux density 0.40 T. Calculate the flux linkage when the normal to the coil is parallel to the field, and state its value when the normal is at 90° to the field.
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Normal parallel: NΦ = BAN = 0.40 × 0.02 × 50 (1)
NΦ = 0.40 Wb-turns (1)
Normal at 90°: NΦ = BAN cos 90° = 0 (1)A coil of 500 turns and area 0.008 m2 lies with its plane perpendicular to a magnetic field. The flux density increases from 0.10 T to 0.30 T. Calculate the change in flux linkage of the coil.
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Δ(NΦ) = NAΔB (1)
= 500 × 0.008 × (0.30 − 0.10) (1)
Δ(NΦ) = 0.80 Wb-turns (1)Describe how a search coil can be used to measure the flux density of a magnetic field.
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A small coil of known area and number of turns is connected to an oscilloscope (or data logger) and placed with its plane perpendicular to the field (1). The flux through it is changed, for example by pulling it out of the field quickly or by using an alternating field, inducing an emf (1). The induced emf equals the rate of change of flux linkage, so B can be calculated from the measured emf, N and A (1).A student must choose a search coil to detect a weak alternating magnetic field. For a given field, the amplitude of the oscilloscope trace is proportional to the peak flux linkage of the coil. Two coils are available.
Coil P: 500 turns, area 1.5 cm2
Coil Q: 2000 turns, area 0.80 cm2
Deduce which coil the student should choose.Mark scheme
For a given field the flux linkage is BANcosθ, so the trace amplitude is proportional to NA (1)
P: NA = 500 × 1.5 × 10−4 = 0.075 m2-turns (1)
Q: NA = 2000 × 0.80 × 10−4 = 0.16 m2-turns (1)
Q gives a trace about twice as tall as P, so the student should choose Q (1)A student uses a search coil connected to an oscilloscope to investigate how the flux linkage of the coil depends on its orientation in an alternating magnetic field. Describe how the student should carry out the investigation and how the results confirm that the flux linkage is proportional to cosθ, where θ is the angle between the field and the normal to the coil.
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Place the search coil in the alternating field, connected to the oscilloscope, starting face-on to the field (1). The trace amplitude is proportional to the peak flux linkage only while the field's angular frequency and amplitude are unchanged, since the peak induced emf scales with both, so the supply settings must be left alone throughout (1). Rotate the coil in measured steps, recording the angle θ between the normal and the field, for example every 15° from 0 to 90° (1). Read the trace amplitude at each angle, keeping the coil at the same position in the field so it samples the same field amplitude (1). Plot the amplitude against cosθ (1). A straight line through the origin confirms that the flux linkage is proportional to cosθ (1).
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