Question

A wheel of 8 metallic spokes each 50cm long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of Earth's magnetic field BH at a place. If BH=0.4G at the place, what is the induced EMF between the axle and the rim of the wheel?

Answer 1

Dear student,
$$\begin{gathered} Verticalcomponentofearthmagneticfieldisgivenby: \\ {B}_v=B\sin \delta \\ B=0.4G \\ \delta =60 \\ {B}_v=B\sin 60 \\ {B}_v=0.4\times \frac{\sqrt{3}}{2} \\ {B}_v=0.692G=0.692\times {10}^{-4}tesla \\ lengthofthestrokewillactasthediameter. \\ radiusr=\frac{50}{2} \\ r=25cm=0.25m \\ Areaa=\pi r^2 \\ a=3.14\times 0.25\times 0.25 \\ a=0.19m^2 \end{gathered}$$
$$\begin{gathered} \omega =2\pi \nu \\ \nu =120revolutionpermin=2revpersecond \\ \omega =2\times 3.14\times 2 \\ \omega =12.56radpersecond \\ inducedemfisonestoke \\ e=B_{v}a\omega \cos \omega t \\ e=0.692\times {10}^{-4}\times 0.19\times 12.56\cos \omega t \\ e=0.1\times {10}^{-3}volt=0.1millivolt \\ forthe8stokes \\ e=8\times 0.1millivolt \\ e=0.8millivolt \\ asnoofstokesincreases. \\ inducedemgincreases. \end{gathered}$$
Regards