Two particles each of mass m and charge q are | vedclass

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

Question

(A) The current $i$ produced by the rotating charge $q$ is given by $i = \frac{q}{T} = \frac{q}{2\pi / \omega} = \frac{q\omega}{2\pi}$.The magnetic mo

Vedclass · Accepted Answer

$\frac{q}{2m}$

Vedclass · Answer

(A) The current $i$ produced by the rotating charge $q$ is given by $i = \frac{q}{T} = \frac{q}{2\pi / \omega} = \frac{q\omega}{2\pi}$.The magnetic moment $M$ of the system is $M = iA$,where $A = \pi R^2$ is the area of the circle traced by the particles.$M = \left( \frac{q\omega}{2\pi} \right) (\pi R^2) = \frac{q\omega R^2}{2}$.The angular momentum $L$ of the system about the centre is $L = I\omega$,where $I$ is the moment of inertia.For two particles of mass $m$ at distance $R$ from the axis,$I = mR^2 + mR^2 = 2mR^2$.Thus,$L = (2mR^2)\omega = 2mR^2\omega$.The ratio of the magnitudes is $\frac{M}{L} = \frac{q\omega R^2 / 2}{2mR^2\omega} = \frac{q}{4m}$.Wait,re-evaluating the current: Since there are two charges,the total current $i = \frac{2q}{T} = \frac{2q\omega}{2\pi} = \frac{q\omega}{\pi}$.Then $M = iA = (\frac{q\omega}{\pi})(\pi R^2) = q\omega R^2$.Then $\frac{M}{L} = \frac{q\omega R^2}{2mR^2\omega} = \frac{q}{2m}$.

Similar Questions

Two wires of the same length are used to form a circle and a square. They carry the same current. The ratio of the magnetic moment of the circle to that of the square is:

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 ( $\mu_0 = \text{permeability of air}$ ).

$A$ wire of length $2\,m$ is bent into a circular loop. When a current of $1\,A$ is passed through the loop,then the magnetic moment of the loop is

The magnetic dipole moment of a rectangular current-carrying loop is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module