ABSTRACT
Problem 2. At what velocity must a conductor 75 mm long cut a magnetic field of flux density 0.6 T if an e.m.f. of 9 V is to be induced in it? Assume the conductor, the field and the direction of motion are mutually perpendicular
Induced e.m.f. E = Blv, hence velocity v = E Bl
Hence velocity
v = 9(0.6)(75 × 10−3)
= 9 × 10 3
0.6 × 75 = 200 m/s
Problem 3. A conductor moves with a velocity of 15 m/s at an angle of (a) 90◦ , (b) 60◦ and (c) 30◦ to a magnetic field produced between two squarefaced poles of side length 2 cm. If the flux leaving a pole face is 5 μWb, find the magnitude of the induced e.m.f. in each case
v = 15 m/s, length of conductor in magnetic field l = 2 cm = 0.02 m, A = 2 × 2 cm2 = 4 × 10−4 m2 and = 5 × 10−6 Wb. (a) E90 = Blv sin 90◦ =
(
A
) lv sin 90◦
= (
5 × 10−6 4 × 10−4
) (0.02)(15)(1) = 3.75 mV
(b) E60 = Blv sin 60◦ = E90 sin 60◦
(c) E30 = Blv sin 30◦ 30◦
= 3.75 sin 30◦ mV
Problem 4. The wing span of a metal aeroplane is 36 m. If the aeroplane is flying at 400 km/h, determine the e.m.f. induced between its wing tips. Assume the vertical component of the Earth’s magnetic field is 40 μT
Induced e.m.f. across wing tips, E = Blv,
B = 40 μT = 40 × 10−6 T, l = 36 m
and v = 400 km h
× 1000 m km
× 1 h 60 × 60 s
= (400)(1000) 3600
= 4000 36
m/s. Hence,
E = Blv = (40 × 10−6)(36) (
4000 36
) = 0.16 V
Problem 5. The diagram shown in Figure 39.4 represents the generation of e.m.f.s. Determine (a) the direction in which the conductor has to be moved in Figure 39.4(a), (b) the direction of the induced e.m.f. in Figure 39.4(b), (c) the polarity of the magnetic system in Figure 39.4(c)
The direction of the e.m.f., and thus the current due to the e.m.f. may be obtained by either Lenz’s law or Fleming’s right-hand rule (i.e. GeneRator rule). (a) Using Lenz’s law: The field due to the magnet and
the field due to the current-carrying conductor are shown in Figure 39.5(a) and are seen to reinforce to the left of the conductor. Hence the force on the conductor is to the right. However, Lenz’s law states that the direction of the induced e.m.f. is
Thus the the left.
