Question

a wheel with 15 metallic spokes each of 60 cm long is rotated at 360 rev/min in a plane normal to the horizontal component of the earth's magnetic field. The angle of dip at that place is 60. If the emf induced between rim of the wheel and thr axle is 400 mV calculate the horizontal component of earth's field at the place.

Answer 1

$$\begin{gathered} Dearstudent, \\ emfinducedinthiscaseisgivenby, \\ \in =B{l}^2\mathrm{\pi v} \\ \mathrm{B}=\frac{\in }{{\mathrm{l}}^2\mathrm{\pi v}} \\ \mathrm{B}=\frac{400\times {10}^{-3}}{(0.60{)}^2\times 3.14\times \left({\displaystyle \frac{360}{60}}\right)} \\ \mathrm{B}=0.058\mathrm{T} \\ \mathrm{for}\mathrm{Earth}\text{'}\mathrm{s}\mathrm{magnetic}\mathrm{field}, \\ \mathrm{Horizontal}\mathrm{component}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{B}}_{\mathrm{H}}=\left|\mathrm{B}\right|\mathrm{Cos\delta } \\ {\mathrm{B}}_{\mathrm{H}}=0.058\times \mathrm{Cos}{60}^{\mathrm{o}} \\ {\mathrm{B}}_{\mathrm{H}}=0.0294\mathrm{T} \\ \mathrm{Regards} \end{gathered}$$