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A wheel with $$10$$  metallic spokes each $$0.5\ \text{m}$$ long is rotated with a speed of   $$120\; \text{rev/min}$$ in a plane normal to the horizontal component of earth’s magnetic field $$H_E$$  at a place. If $$H_E=0.4\;G$$  at the place, what is the induced emf between the axle and the rim of the wheel? Note that $$1\;G=10^{-4}\;T$$ .
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## QuestionPhysicsClass 12

A wheel with $$10$$  metallic spokes each $$0.5\ \text{m}$$ long is rotated with a speed of   $$120\; \text{rev/min}$$ in a plane normal to the horizontal component of earth’s magnetic field $$H_E$$  at a place. If $$H_E=0.4\;G$$  at the place, what is the induced emf between the axle and the rim of the wheel? Note that $$1\;G=10^{-4}\;T$$ .

Given that,
Angular speed of Wheel $$\left(\omega \right)=120\;\frac{\mathrm{rev}}{\mathrm{min}}=120\times \frac{\pi } 3=4\pi \;\mathrm{rad}/\mathrm{sec}$$
Radius of the wheel $$\left(R\right)=0.5\;\text{m}$$
Horizontal magnetic field $$(H)_E\mid B)=0.4\;G$$
Let us consider a small elemental length $$^\prime \mathit{dl}^\prime$$  of the metallic spoke at a distance $$l$$  from the axle $$($$ centre of the wheel to considered as the origin $$)$$ moving at a linear speed $$v$$  in a magnetic field $$B$$  perpendicular to the length and speed.
The emf induced across the elemental length $$\mathit{dl}$$
$${\therefore}$$   $$\mathit{d\varepsilon }=\mathit{vBdl}$$
Since the wheel is rotating at an angular speed $$\omega$$ , we have
$${\therefore}v=\mathit{\omega l}$$
$$\Rightarrow \mathit{d\varepsilon }=\mathit{\omega lBdl}$$
The total emf induced across the metallic spoke is given by:
$${\therefore}\varepsilon =\int _0^R\mathit{d\varepsilon }=\int _0^R\mathit{\omega lBdl}$$
$$=\mathit{\omega B}\int _0^R\mathit{ldl}$$
$$=\frac 1 2\mathit{\omega B}R^2$$
$$=\frac 1 2\times 4\pi \times 0.4\times 10^{-4}\times (0.5)^2$$
$$\Rightarrow \varepsilon =6.28\times 10^{-5}V$$
Since all the metallic spokes are parallel as one end is connected to common axle and other end is connected to common rim. So emf across all the metallic spokes will be the same.