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A cycle wheel of radius $0.5 \mathrm{~m}$ is rotated with constant angular velocity of $10 \mathrm{rad} / \mathrm{s}$ in a region of magnetic field of $0.1 \mathrm{~T}$ which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is
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$0.125 \mathrm{~V}$
When a conducting disc or wheel of radius $r$ rotates with constant angular velocity of $\omega$ about its axis in a uniform magnetic field perpendicular to its plane and parallel to its axis of rotation, then,
Induced emf is given by, $\mathrm{e}=\frac{1}{2}{\mathrm{~B} \omega \mathrm{r}^2}^2$
Here, $B=0.1 \mathrm{~T}, \omega=10 \mathrm{rad} / \mathrm{s}, \mathrm{r}=0.5 \mathrm{~m}$
Substituting these values in Eq. (i), we get
$$
\therefore \quad \mathrm{e}=\frac{1}{2} \times 0.1 \times 10 \times(0.5)^2=\frac{1}{8} \mathrm{~V}=0.125 \mathrm{~V}
$$
Induced emf is given by, $\mathrm{e}=\frac{1}{2}{\mathrm{~B} \omega \mathrm{r}^2}^2$
Here, $B=0.1 \mathrm{~T}, \omega=10 \mathrm{rad} / \mathrm{s}, \mathrm{r}=0.5 \mathrm{~m}$
Substituting these values in Eq. (i), we get
$$
\therefore \quad \mathrm{e}=\frac{1}{2} \times 0.1 \times 10 \times(0.5)^2=\frac{1}{8} \mathrm{~V}=0.125 \mathrm{~V}
$$
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