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What is electromotive force induced between its ends

1 vote

Consider a quarter circular conducting ring of large radius r with its
centre at the origin, where a magnetic dipole of moment m is placed as
shown in the figure. If the ring rotates at a constant angular velocity $\omega$
about the y-axis, then what is electromotive force induced between its ends?

My try :

asked Aug 18, 2018 in Physics Problems by koolman (4,156 points)
You seem to have changed the axes from those given in the question.

You state that the direction of $B$ from the dipole should be "along the x axis" (ie vertical). It will have a component along the y axis also (ie horizontal), except for points along the x (vertical) axis. Is there a component of $B$ along the z axis, ie perpendicular to the plane in which the quarter ring lies?

Your "circuit" appears to be imaginary conducting rods OA, OB and the quarter ring. This circuit is in the xy plane. If there is no component of the magnetic field perpendicular to this plane (z direction), how can there be any flux through the circuit?

1 Answer

1 vote
Best answer

You should expect the EMF induced in the quarter circle to be constant because of the symmetry of the magnetic field of the dipole : whatever angle the quarter circle turns through, the magnetic field distribution in the plane of the quarter circle always looks the same.

Imagine that the quarter circle is located on the surface of a sphere of radius $r$. As it rotates about the vertical axis of the sphere, it sweeps out a region $S$ on the surface of the sphere. After one complete revolution $S$ is the surface of a hemisphere.

The magnetic field due to the magnetic dipole has radial, tangential and azimuthal components. But we are only interested in the radial component $B(r)$ which crosses the surface of the sphere. The flux of $B(r)$ across the upper hemisphere is the flux through the area swept out by the quarter ring as it makes one complete revolution. This is the flux $\phi$ which you need to use in Faraday's Law $E=-\frac{d\phi}{dt}$.

The radial component of the magnetic field of the dipole is $$B(r)=\frac{\mu_0 m}{4\pi r^3} (2\cos\theta)$$ where $m$ is the magnitude of the dipole moment $\mathbf{m}$ and $\theta$ is the angle between the radius and the dipole axis. (See eg wiki: Magnetic Dipole External Field.)

The area of an elementary hoop of the sphere of radius $R=r\sin\theta$ and width $rd\theta$ at polar angle $\theta$ is $$dA=2\pi R. rd\theta=2\pi r^2 \sin\theta d\theta$$ So the flux through the upper hemisphere is $$\Phi=\int_0^{\pi/2} B(r)dA =\frac{\mu_0 m}{2r}
\int_0^{\pi/2} \sin2\theta d\theta=\frac{\mu_0 m}{2r}$$

The rate at which this flux is cut by the quarter circle equals the EMF induced in the ring : $$E=\frac{d\phi}{dt}=\frac{\omega}{2\pi}\Phi=\frac{\mu_0 m\omega}{4\pi r}$$ because $$\phi=\frac{t}{T}\Phi=\frac{\omega}{2\pi}\Phi t$$ where $T$ is the time for the quarter circle to make one rotation.

answered Aug 18, 2018 by sammy gerbil (26,660 points)
selected Aug 22, 2018 by koolman