Q#6 (Past Exam Paper – June 2008 Paper 4 Q6)
A small rectangular coil ABCD contains 140 turns of wire. The sides AB and BC of the coil are of lengths 4.5 cm and 2.8 cm respectively, as shown in Fig. 6.1.
Fig. 6.1
The coil is held between the poles of a large magnet so that the coil can rotate about an axis through its centre.
The magnet produces a uniform magnetic field of flux density B between its poles.
When the current in the coil is 170 mA, the maximum torque produced in the coil is
2.1 × 10-3 N m.
(a) For the coil in the position for maximum torque, state whether the plane of the coil is
parallel to, or normal to, the direction of the magnetic field. [1]
(b) For the coil in the position shown in Fig. 6.1, calculate the magnitude of the force on
(i) side AB of the coil, [2]
(ii) side BC of the coil. [1]
(c) Use your answer to (b)(i) to show that the magnetic flux density B between the poles of the magnet is 70 mT. [2]
(d) (i) State Faraday’s law of electromagnetic induction. [2]
(ii) The current in the coil in (a) is switched off and the coil is positioned as shown in
Fig. 6.1.
The coil is then turned through an angle of 90° in a time of 0.14 s.
Calculate the average e.m.f. induced in the coil. [3]
Solution:
(a) parallel (to the field)
{When the plane of the coil is horizontal (as in the diagram), there is maximum flux (cutting of the field).}
(b)
(i)
torque = F × d
{The force and distance should be perpendicular.}
2.1×10-3 = F × 2.8×10-2
F = 0.075 N
(ii) zero
{The direction of current in BC is parallel to the magnetic field. So, no force is induced.}
(c)
F = BILN (sinθ)
F = BILN (sinθ)
{The angle between the force and the field is 90°. So, sin 90° = 1.
Also since there are 140 turns of wires, we need to include the number of turns in the formula.}
0.075 = B × 0.170 × 4.5×10-2 × 140
B = 7.0×10-2 T = 70 mT
(d) (i) Faraday’s law of electromagnetic induction states that the (induced) e.m.f. is proportional to the rate of change of (magnetic) flux (linkage).
(ii)
change in flux linkage ϕ = BAN
ϕ = 0.070 × 4.5×10-2 × 2.8×10-2 × 140
ϕ = 0.0123 Wb turns
{induced e.m.f. = Δϕ / Δt}
induced e.m.f = 0.0123 / 0.14
induced e.m.f = 88 mV
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