Question 6
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]
Reference: Past Exam Paper – June 2008 Paper 4 Q6
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 Ï• = BAN
Ï• = 0.070 × 4.5×10-2 × 2.8×10-2 × 140
Ï• = 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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