EMI O4

1. Problem Setup and Motional EMF Formula

We have a quarter-circular conducting ring of radius $r$ in the $xy$-plane. A magnetic dipole with moment $\vec{m}$ is placed at the origin, oriented along the $y$-axis ($\vec{m} = m\hat{j}$). The ring rotates with a constant angular velocity $\omega$ about the $y$-axis. We need to find the electromotive force (EMF) induced between the ends of the ring, A and B.

The induced EMF in a moving conductor is given by the line integral of the motional electric field $\vec{E}_m = \vec{v} \times \vec{B}$ along the conductor:

2. Velocity and Magnetic Field Vectors

Consider an arbitrary point on the ring at an angle $\theta$ from the x-axis. Its position is $\vec{r}’ = r(\cos\theta \hat{i} + \sin\theta \hat{j})$.

• Velocity ($\vec{v}$): The ring rotates about the y-axis. The velocity of a point at distance $x = r\cos\theta$ from the rotation axis is in the $-\hat{k}$ direction (into the page). $$ \vec{v} = \vec{\omega} \times \vec{r}’ = (\omega \hat{j}) \times (r\cos\theta \hat{i} + r\sin\theta \hat{j}) = -\omega r \cos\theta \hat{k} $$
• Magnetic Field ($\vec{B}$): The magnetic field of a dipole $\vec{m}$ at a position $\vec{r}’$ is given by: $$ \vec{B}(\vec{r}’) = \frac{\mu_0}{4\pi r’^3} [3(\vec{m} \cdot \hat{r}’)\hat{r}’ – \vec{m}] $$ With $\vec{m} = m\hat{j}$ and $\hat{r}’ = \cos\theta \hat{i} + \sin\theta \hat{j}$, we have $\vec{m} \cdot \hat{r}’ = m\sin\theta$. The field is in the $xy$-plane.

3. Calculation of the EMF

We can use the vector identity $(\vec{v} \times \vec{B}) \cdot d\vec{l} = \vec{B} \cdot (d\vec{l} \times \vec{v})$. The vector $d\vec{l}$ is along the ring, and $\vec{v}$ is perpendicular to the plane of the ring. Thus, $d\vec{l} \times \vec{v}$ is a vector in the $xy$-plane, perpendicular to the ring (radial direction). $$ d\vec{l} \times \vec{v} = (dx\hat{i} + dy\hat{j}) \times (-\omega x \hat{k}) = \omega x dy \hat{i} – \omega x dx \hat{j} $$ Using polar coordinates ($x=r\cos\theta, y=r\sin\theta$) on the ring, $dx = -r\sin\theta d\theta$ and $dy = r\cos\theta d\theta$. $$ d\vec{l} \times \vec{v} = \omega(r\cos\theta)(r\cos\theta d\theta)\hat{i} – \omega(r\cos\theta)(-r\sin\theta d\theta)\hat{j} $$ $$ d\vec{l} \times \vec{v} = \omega r^2 \cos\theta d\theta (\cos\theta \hat{i} + \sin\theta \hat{j}) = \omega r^2 \cos\theta d\theta \hat{r}’ $$ Now, we need the radial component of the magnetic field, $B_r = \vec{B} \cdot \hat{r}’$. $$ B_r = \frac{\mu_0}{4\pi r^3} [3(m\sin\theta)(\hat{r}’\cdot\hat{r}’) – (m\hat{j}\cdot\hat{r}’)] = \frac{\mu_0}{4\pi r^3} [3m\sin\theta – m\sin\theta] = \frac{2\mu_0 m \sin\theta}{4\pi r^3} = \frac{\mu_0 m \sin\theta}{2\pi r^3} $$ Integrating from A ($\theta=\pi/2$) to B ($\theta=0$): $$ \varepsilon = \int_{\pi/2}^{0} \vec{B} \cdot (d\vec{l} \times \vec{v}) = \int_{\pi/2}^{0} B_r (\omega r^2 \cos\theta d\theta) $$ $$ \varepsilon = \int_{\pi/2}^{0} \left( \frac{\mu_0 m \sin\theta}{2\pi r^3} \right) (\omega r^2 \cos\theta) d\theta = \frac{\mu_0 m \omega}{2\pi r} \int_{\pi/2}^{0} \sin\theta \cos\theta d\theta $$ Using $\sin\theta\cos\theta = \frac{1}{2}\sin(2\theta)$: $$ \varepsilon = \frac{\mu_0 m \omega}{4\pi r} \int_{\pi/2}^{0} \sin(2\theta) d\theta = \frac{\mu_0 m \omega}{4\pi r} \left[ -\frac{\cos(2\theta)}{2} \right]_{\pi/2}^{0} $$ $$ \varepsilon = \frac{\mu_0 m \omega}{8\pi r} [-\cos(0) – (-\cos(\pi))] = \frac{\mu_0 m \omega}{8\pi r} [-1 – (-(-1))] = \frac{\mu_0 m \omega}{8\pi r} (-2) = -\frac{\mu_0 m \omega}{4\pi r} $$ The magnitude of the induced EMF is:

4. Conclusion

The magnitude of the electromotive force induced between the ends of the quarter-circular ring is $\frac{\mu_0 m \omega}{4\pi r}$.