A thin circular disk of radius R is uniformly | vedclass

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

Question

(C) Consider a thin elemental ring of radius $x$ and width $dx$ on the disk.The area of this elemental ring is $dA = 2 \pi x dx$.The charge on this el

Vedclass · Accepted Answer

$\frac{1}{4} \pi R^{4} \sigma \omega$

Vedclass · Answer

(C) Consider a thin elemental ring of radius $x$ and width $dx$ on the disk.The area of this elemental ring is $dA = 2 \pi x dx$.The charge on this elemental ring is $dq = \sigma dA = 2 \pi \sigma x dx$.Since the disk rotates with angular speed $\omega$,the time period of rotation is $T = \frac{2 \pi}{\omega}$.The equivalent current $dI$ due to this rotating charge is $dI = \frac{dq}{T} = \frac{dq \cdot \omega}{2 \pi}$.Substituting $dq$,we get $dI = \frac{(2 \pi \sigma x dx) \omega}{2 \pi} = \sigma \omega x dx$.The magnetic moment $dM$ of this elemental ring is $dM = dI \cdot A_{ring} = (\sigma \omega x dx) \cdot (\pi x^2) = \pi \sigma \omega x^3 dx$.To find the total magnetic moment $M$,integrate $dM$ from $x = 0$ to $x = R$:$M = \int_{0}^{R} \pi \sigma \omega x^3 dx = \pi \sigma \omega \left[ \frac{x^4}{4} \right]_{0}^{R} = \frac{1}{4} \pi \sigma \omega R^4$.

$A$ thin circular disk of radius $R$ is uniformly charged with surface charge density $\sigma > 0$. The disk rotates about its central axis with a uniform angular speed $\omega$. The magnetic moment of the disk is:

Similar Questions

The magnetic field at the centre of a current carrying circular coil of an area $A$ is $B$. The magnetic moment of the coil is (where $\mu_0$ is the permeability of free space).

Magnetic induction at the centre of a circular loop of area $\pi \ m^2$ is $0.1 \ T$. The magnetic moment of the loop is ( $\mu_0 = \text{permeability of air}$ ).

An insulating thin rod of length $l$ has a linear charge density $\rho(x) = \rho_0 \frac{x}{l}$ on it. The rod is rotated about an axis passing through the origin $(x = 0)$ and perpendicular to the rod. If the rod makes $n$ rotations per second,then the time-averaged magnetic moment of the rod is:

Derive an expression for the torque acting on a current-carrying loop suspended in a uniform magnetic field.

Vedclass Test Series

Exam Paper Generator

Online Exam Module