(D) The induced emf $(e)$ in a rotating metallic spoke of length $(l)$ in a magnetic field $(B)$ with angular velocity $(\omega)$ is given by the form

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(D) The induced emf $(e)$ in a rotating metallic spoke of length $(l)$ in a magnetic field $(B)$ with angular velocity $(\omega)$ is given by the formula: $e = \frac{1}{2} B \omega l^2$.Here, the emf is independent of the number of spokes, as each spoke acts as an individual source of emf connected in parallel between the axle and the rim.Given the initial state: $\omega_1 = 180 \ rev/min$, $e_1 = E = \frac{1}{2} B \omega_1 l^2$.Given the final state: $\omega_2 = 90 \ rev/min$, $e_2 = \frac{1}{2} B \omega_2 l^2$.Taking the ratio: $\frac{e_2}{E} = \frac{\omega_2}{\omega_1} = \frac{90}{180} = \frac{1}{2}$.Therefore, $e_2 = 0.5 E$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

Medium

If a wheel with $24$ metallic spokes each $40 \ cm$ long is rotated with a speed of $180 \ rev/min$ in a plane normal to the horizontal component of earth's magnetic field, the emf induced between the axle and the rim of the wheel is $E$. If the number of spokes is made $12$ and the wheel is rotated with a speed of $90 \ rev/min$ in the same field, the induced emf is

AP EAMCET 2025 All AP EAMCET

A
$E$
B
$2E$
C
$4E$
D
$0.25 E$

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

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Consider the situation shown in the figure. The wire $AB$ is sliding on the fixed rails with a constant velocity. If the wire $AB$ is replaced by a semicircular wire,the magnitude of the induced current will

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