Question 32: Electromagnetic Induction

Question

A metal rod of length $L$ rotates about one end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(\mathrm{r})=B_{\mathrm{o}} \mathrm{e}^{-\lambda r} ; \lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

Accepted Answer

Let a small element of length $d r$ be located at a distance $r$ from the axis of rotation (the origin). The velocity $v$ of this element is given by :
$$ \mathrm{v}=\mathrm{r} \omega $$
The magnetic field at distance r is $\mathrm{B}(\mathrm{r})=\mathrm{B}_0 \mathrm{e}^{-\lambda \mathrm{r}}$.
The emf $\mathrm{d} \epsilon$ induced across this small element dr moving perpendicular to the magnetic field is:
$$ \mathrm{d} \epsilon=\mathrm{B}(\mathrm{r}) \cdot \mathrm{v} \cdot \mathrm{dr} $$
$$\Rightarrow $$ $d \epsilon=\left(B_0 e^{-\lambda r}\right) \cdot(r \omega) \cdot d r$
$$\Rightarrow $$ $$ \mathrm{d} \epsilon=\mathrm{B}_0 \omega\left(\mathrm{re}^{-\lambda \mathrm{r}}\right) \mathrm{dr} $$
To find the total emf $\epsilon$, we integrate from $r=0$ to $r=L$ :
$$ \epsilon=\int_0^{\mathrm{L}} \mathrm{~B}_0 \omega \mathrm{r} \mathrm{e}^{-\lambda \mathrm{r}} \mathrm{dr} $$
$$\Rightarrow $$ $$ \epsilon=\mathrm{B}_0 \omega \int_0^{\mathrm{L}} \mathrm{re}^{-\lambda \mathrm{r}} \mathrm{dr} $$
Using integration by parts, let $\mathrm{u}=\mathrm{r} \Rightarrow \mathrm{du}=\mathrm{dr}$ Let $d v=e^{-\lambda r} d r \Rightarrow v=\frac{e^{-\lambda r}}{-\lambda}$
$$ \epsilon=\mathrm{B}_0 \omega \int_0^{\mathrm{L}} \mathrm{r} \mathrm{e}^{-\lambda \mathrm{r}} \mathrm{dr} $$
$$\Rightarrow $$ $\epsilon=\mathrm{B}_0 \omega\left[\left[\mathrm{r} \frac{\mathrm{e}^{-\lambda \mathrm{r}}}{-\lambda}\right]_0^{\mathrm{L}}-\int_0^{\mathrm{L}} \frac{\mathrm{e}^{-\lambda \mathrm{r}}}{-\lambda} \mathrm{dr}\right]$
$$\Rightarrow $$ $\epsilon=\mathrm{B}_0 \omega\left[-\frac{\mathrm{Le}^{-\lambda \mathrm{L}}}{\lambda}+\frac{1}{\lambda}\left[\mathrm{e}^{-\lambda \mathrm{r}}-\lambda\right]_0^{\mathrm{L}}\right]$
$$\Rightarrow $$ $\epsilon=\mathrm{B}_0 \omega\left[-\mathrm{L} \frac{\mathrm{e}^{-\lambda \mathrm{L}}}{\lambda}-\frac{1}{\lambda^2}\left[\mathrm{e}^{-\lambda \mathrm{L}}-\mathrm{e}^0\right]\right]$
$$\Rightarrow $$ $$ \epsilon=\mathrm{B}_0 \omega\left[\frac{-\mathrm{Le}^{-\lambda \mathrm{L}}}{\lambda}-\frac{\mathrm{e}^{-\lambda \mathrm{L}}}{\lambda^2}+\frac{1}{\lambda^2}\right] $$
Therefore, the induced emf is $B_o \omega\left[\frac{1}{\lambda^2}-e^{-\lambda L}\left(\frac{1}{\lambda^2}+\frac{L}{\lambda}\right)\right]$.
Hence, the correct option is (A).

Answer

$$ B_o \omega\left[\frac{1}{\lambda^2}-e^{-\lambda L}\left(\frac{1}{\lambda^2}+\frac{L}{\lambda}\right)\right] $$

Answer

$$ B_o \omega\left[\frac{1}{\lambda^2}+e^{-\lambda L}\left(\frac{1}{\lambda^2}+\frac{L}{\lambda}\right)\right] $$

Answer

$$ B_o \omega\left[\frac{4}{\lambda^2}-e^{-2 \lambda L}\left(\frac{1}{\lambda^2}+\frac{2 L}{\lambda}\right)\right] $$

Answer

$$ B_0 \omega\left[\frac{3}{\lambda^2}-e^{-3 \lambda L}\left(\frac{3}{\lambda^2}+\frac{L}{\lambda}\right)\right] $$

Electromagnetic Induction

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