MHT CET · Physics · Electromagnetic Induction
A bicycle wheel of radius ' \(R\) ' has ' \(n\) ' spokes. It is rotating at the rate of ' \(F\) ' r.p.m. perpendicular to the horizontal component of earth's magnetic field \(\vec{B}\). The e.m.f. induced between the rim and the centre of the wheel is
- A \(\frac{1}{2}{\mathrm{~B} \pi \mathrm{FR}^2}\)
- B \(\quad \mathrm{B} \pi \mathrm{FR}^2\)
- C \(\frac{1}{\mathrm{n}} \mathrm{B} \pi \mathrm{FR}\)
- D \(\mathrm{B} \pi \mathrm{FR}^2 \mathrm{n}\)
Answer & Solution
Correct Answer
(B) \(\quad \mathrm{B} \pi \mathrm{FR}^2\)
Step-by-step Solution
Detailed explanation
As the angular velocity is constant, the linear velocity changes from 0 to \(R \omega\) as we move from the axle to the rim.
\(\therefore \quad\) Average velocity \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{R} \omega}{2}\)
We know motional e.m.f, \(\mathrm{E}_{\mathrm{mot}}=\mathrm{B} / \mathrm{v}\)
\(\begin{aligned}
\therefore \quad \mathrm{E}_{\mathrm{mot}} & =\frac{\mathrm{B} \times \mathrm{R} \times \mathrm{R} \omega}{2}=\frac{\mathrm{BR}^2 \times 2 \pi \mathrm{~F}}{2} \\
& =\mathrm{B} \pi \mathrm{FR}^2
\end{aligned}\)
\(\therefore \quad\) Average velocity \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{R} \omega}{2}\)
We know motional e.m.f, \(\mathrm{E}_{\mathrm{mot}}=\mathrm{B} / \mathrm{v}\)
\(\begin{aligned}
\therefore \quad \mathrm{E}_{\mathrm{mot}} & =\frac{\mathrm{B} \times \mathrm{R} \times \mathrm{R} \omega}{2}=\frac{\mathrm{BR}^2 \times 2 \pi \mathrm{~F}}{2} \\
& =\mathrm{B} \pi \mathrm{FR}^2
\end{aligned}\)
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