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

$A$ uniformly charged dielectric ring rotating with constant angular velocity $\omega$ about its geometrical axis produces:

$A$ straight wire carrying current $i$ is turned into a circular loop. If the magnitude of the magnetic moment associated with it in $MKS$ units is $M$,the length of the wire will be

Two concentric circular loops of radii $r_{1} = 30 \, cm$ and $r_{2} = 50 \, cm$ are placed in the $X-Y$ plane as shown in the figure. $A$ current $I = 7 \, A$ flows through them in the directions shown. The net magnetic moment of this system of two circular loops is approximately:

An electron revolving in a circular orbit of radius $r$ with velocity $v$ and frequency $f$ has an orbital magnetic moment $M$. If the frequency of revolution is doubled,then the new magnetic moment will be:

$A$ thin circular wire carrying a current $I$ has a magnetic moment $M$. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment

Vedclass Test Series

Exam Paper Generator

Online Exam Module