Q#5 (Past Exam Paper – June 2018 Paper 42 Q9)
(a) State what is meant by the magnetic flux linkage of a coil. [3]
(b) A coil of wire has 160 turns and diameter 2.4 cm. The coil is situated in a uniform magnetic field of flux density 7.5 mT, as shown in Fig. 9.1.
Fig. 9.1
The direction of the magnetic field is along the axis of the coil.
The magnetic flux density is reduced to zero in a time of 0.15 s.
Show that the average e.m.f. induced in the coil is 3.6 mV. [2]
(c) The magnetic flux density B in the coil in (b) is now varied with time t as shown in Fig. 9.2.
Fig. 9.2
Use data in (b) to show, on Fig. 9.3, the variation with time t of the e.m.f. E induced in the coil.
Fig. 9.3
[4]
[Total: 9]
Solution:
(a)
{ϕ = NBA or ϕ = NBA sinθ where θ is the angle between B and A
Magnetic flux linkage of a coil is the product of the magnetic flux density (B) with the cross-sectional area (A) {normal to the magnetic flux density} and the number of turns (N) on the coil.}
(b)
e.m.f. = BAN / t
or
e.m.f = rate of change of flux linkage
{N = 160 turns
B = 7.5×10-3 T
Diameter = 2.4 cm so, radius r = 1.2 cm = 1.2×10-2 m
Area A = π r2}
e.m.f. = (7.5×10-3 × π × {1.2×10-2}2 × 160) / 0.15
e.m.f. = 3.6 × 10-3 V
(c)
sketch: zero for 0–0.10 s, 0.25–0.35 s, and 0.425–0.55 s, and non-zero outside these ranges
two horizontal steps, with zero voltage either side
with same polarity
correct values (1st step 3.6 mV and 2nd step 7.2 mV)
When an emf is induced?:
An e.m.f. is induced when there is a change in the flux linkage with time.
If flux linkage is not changing (it is constant – in the graph, this is a horizontal line), no e.m.f. is induced.
So, E = 0 for the times: 0–0.10 s, 0.25–0.35 s, and 0.425–0.55 s
Outside of these range, the e.m.f. is not zero. In the graph, these correspond to horizontal steps.
Polarity of the steps:
An emf is induced when there is a change in flux linkage B.
If B increases, the emf induced would have a polarity (+ or -) and if B decreases, the emf induced would have the opposite polarity.
Let’s say that when B decreases, the polarity of the emf induced is positive. {Note that this is arbitrary – one could also have chosen the polarity to be negative for a decrease in B}.
From the graph, it can be observed that the change in B is a decrease in both cases where it is changing. So, in or case, the emf induced will be positive in each case.
Also, both sloping parts of the B-t graph had the same sign of gradient, so the two steps should each show an e.m.f. of the same polarity.
Magnitude of induced emf:
emf = rate of change of flux linkage
In both cases, the change in B is a decrease by 7.5 mT.
The first decrease occurs from time = 0.1 s to t = 0.25 s. That is, in 0.15 s.
As calculated in part (b), the emf induced during this change is 3.6 mT.
The second decrease occurs from time = 0.35 s to t = 0.425 s. That is, in 0.75 s. This is half the time of the first change.
The same change of magnetic flux linkage occurs over half the time of the first one. This will result in an induced e.m.f. of twice the magnitude of the first.
Here, emf induced = 2 × 3.6 = 7.2 mT
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