Rotating charged half ring

Electricity and Magnetism Level 5

A charged half ring of radius $R$ is held fixed on a plane. Let the center of the half ring be $O$ .

Let us complete the circle $C$ which contains the half ring. A point $P$ is taken on the circumference of $C$ , such that $\overrightarrow {PO}$ passes through the center of mass of the half ring.

The plane is then rotated with an angular velocity $\omega$ about an axis passing through $P$ and perpendicular to the plane.

Find the value of $\dfrac {|\vec{B_P}|}{|\vec{E_P}|} \cdot 10^{16}$ where $|\vec{B_P}|$ and $|\vec{E_P}|$ are the magnitudes of magnetic and electric fields at point $P$ respectively.

$\textbf{Assumptions}$

$\bullet \ \ \$ The mass distribution of the half ring is uniform.

$\bullet \ \ \ \$ The charge distribution of the half ring is uniform.

$\bullet \ \ \ \$ The above procedure is carried out in vacuum, so the permeability $\mu_{o}=4\pi \times 10^{-7} \ \text {H/m}$

$\bullet \ \ \ \$ When the plane rotates, the angular velocity of the half ring with respect to the plane is $0$

$\bullet \ \ \ \$ Coulomb's constant $k=9 \times 10^9 \ \text {N} \text{m}^2/\text {C}^2$

$\bullet \ \ \ \ \omega=9 \ \text{rad/s}$

$\bullet \ \ \ \ R=1 \ \text{m}$

The answer is 2.

1 solution

Jatin Yadav
Aug 19, 2014

Using simple trigonometry, we get:

$AP = 2R \cos \dfrac{x}{2}$

$B_{P} = \displaystyle \int_{-\pi/2}^{\pi/2} \dfrac{\frac{\mu_{0} \lambda R \omega dx}{2 \pi}}{2 \times 2R \cos \frac{x}{2}}$

= $\dfrac{\mu_{0} \lambda \omega}{8 \pi } \displaystyle \int_{-\pi/2}^{\pi/2} \dfrac{dx }{\cos \frac{x}{2}}$

In Electric field only vertical component is summed.

$E_{P} = \displaystyle \int_{-\pi/2}^{\pi/2} \dfrac{\lambda R dx \cos \frac{x}{2}}{4 \pi \epsilon_{0} \times 4R^2 \cos^2 \frac{x}{2}}$

$= \displaystyle \dfrac{\lambda }{16 \pi \epsilon_{0}R}\int_{-\pi/2}^{\pi/2} \dfrac{dx }{\cos \frac{x}{2}}$

Hence, $\dfrac{B_{P}}{E_{P}} = \dfrac{ 2 \frac{\mu_{0}}{4 \pi} \omega R}{\frac{1}{4 \pi \epsilon_{0}}}$

Can anyone please explain what the question is saying.....

Tushar Gopalka - 6 years, 9 months ago

@Tushar Gopalka The question asks to find the ratio of magnetic field and electric field at point $P$ , which is as shown ---

Charged Half-Ring

Pratik Shastri - 6 years, 9 months ago

Thanks....

Tushar Gopalka - 6 years, 9 months ago

Isn't there any elegant method . I thought there would be something to it.

Rohit Shah - 6 years, 8 months ago

Awesome solution.....

Tushar Gopalka - 6 years, 9 months ago

Rotating charged half ring

The answer is 2.

1 solution

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