Question

A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth 's magnetic field at a place.. If BH = 0.4 G at the place .What is the induced emf between the axle and the rim of the wheel?

Answer 1

angular speed , ω = 120 rev/min = 120 × 2π/60 rad/s = 4π rad/s

horizontal component of earth's magnetic field , $B_H$ = 0.4G

as we know, 1G = 10^-4 T

so, $B_H$ = 0.4 × 10^-4 T

here, the number of spokes is immaterial. because, each spoke acts as battery connected parallely between two terminals.

now induced emf , E = $\int{Bvdr}$ , where v = ωr

so, emf = $B\omega \int\limits^r_0{r}\,dr$

= Bωr²/2

= 0.4 × 10^-4 × 4π × (0.5)²/2

= 0.4 × 10^-4 × 2π × 0.25

= 6.28 × 10^-5 Volts .

Answer 2

Answer:

we have no of spoken (N)=10 length of each spoke(L)=0.5m magnetic field (B)=0.4×10−4

T=4×10−5

frequency (f)=120rpm=2rps

induced emf between given by

e=N×B×l^2×π×t

=10×4×−5×0.5×0.5×3.14×2v

=6.28×10^−5V