A plastic disc of radius R has a charge q uni | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Consider a small elemental ring of radius $x$ and thickness $dx$ on the disc.The area of the disc is $A = \pi R^2$. The surface charge density is

Vedclass · Accepted Answer

$\frac{\mu_0 q \omega}{2 \pi R}$

Vedclass · Answer

(A) Consider a small elemental ring of radius $x$ and thickness $dx$ on the disc.The area of the disc is $A = \pi R^2$. The surface charge density is $\sigma = \frac{q}{\pi R^2}$.The charge on the elemental ring is $dq = \sigma \cdot (2\pi x dx) = \frac{q}{\pi R^2} \cdot 2\pi x dx = \frac{2q x dx}{R^2}$.When the disc rotates with angular frequency $\omega$,this ring acts as a current loop with current $dI = \frac{dq}{T} = \frac{dq \cdot \omega}{2\pi} = \frac{2q x dx}{R^2} \cdot \frac{\omega}{2\pi} = \frac{q \omega x dx}{\pi R^2}$.The magnetic field at the center due to this ring is $dB = \frac{\mu_0 dI}{2x} = \frac{\mu_0}{2x} \cdot \frac{q \omega x dx}{\pi R^2} = \frac{\mu_0 q \omega}{2 \pi R^2} dx$.Integrating from $x = 0$ to $x = R$:$B = \int_0^R \frac{\mu_0 q \omega}{2 \pi R^2} dx = \frac{\mu_0 q \omega}{2 \pi R^2} [x]_0^R = \frac{\mu_0 q \omega}{2 \pi R}$.

$A$ plastic disc of radius $R$ has a charge $q$ uniformly distributed over its surface. If the disc is rotated at an angular frequency $\omega$ about its axis,the magnetic induction at the center of the disc is:

Similar Questions

Consider the magnetic field produced by a long straight current-carrying wire.

$A$ long conducting wire carrying a current $I$ is bent at $120^{\circ}$ (see figure). The magnetic field $B$ at a point $P$ on the angle bisector of the bend at a distance $d$ from the bend is ($\mu_{0}$ is the permeability of free space):

Which of the following graphs correctly represents the variation of magnetic field $B$ with distance $R$ from a long straight current-carrying conductor?

$AB$ and $CD$ are long straight conductors,separated by a distance $d$,each carrying a current $I$. The magnetic field at the midpoint of $BC$ is

Two identical coils of radius $R$ and number of turns $N$ are placed perpendicular to each other such that they have a common center. The currents through them are $I$ and $I\sqrt{3}$. The resultant intensity of the magnetic field at the center of the coils will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module