An electron of charge e mass m moves in an orbit of radius r with an angular velocity $ω$ . The ratio of the magnetic moment of the orbiting electron to its angular momentum is
e/m
2e/m
e/2m
2m/e

Question

An electron of charge e mass m moves in an orbit of radius r with an angular velocity omega  - The ratio of the magnetic moment of the orbiting electron to its angular momentum ise-m2e-me-2m2m-e

Toppr Admin · Accepted Answer

Angular momentum-mvr-mr2-x3C9-magnetic moment-xA0-i-x3C0-r2-xA0-dqdt-x3C0-r2-xA0-e-f-x3C0-r2 -e-x3C9-2-x3C0-x3C0-r2-xA0-e-x3C9-r22-xA0-MagneticmomentAngularmoment-e2m

An electron of charge e mass m moves in an orbit of radius r with an angular velocity ω . The ratio of the magnetic moment of the orbiting electron to its angular momentum ise/m2e/me/2m2m/e

angular momentum=mvr=mr2ωmagnetic moment =i.π.r2 =(dqdt).π.r2 =e.f.π.r2 =e.(ω2π).π.r2 =e.ω.r22 MagneticmomentAngularmoment=e2m

An electron of charge e mass m moves in an orbit of radius r with an angular velocity $ω$ . The ratio of the magnetic moment of the orbiting electron to its angular momentum is
e/m
2e/m
e/2m
2m/e

angular momentum=mvr
= $m r^{2} ω$
magnetic moment
= $i . π . r^{2}$
= $(\frac{d q}{d t}) . π . r^{2}$
= $e . f . π . r^{2}$ = $e . (\frac{ω}{2 π}) . π . r^{2}$
= $\frac{e . ω . r^{2}}{2}$
$\frac{M a g n e t i c m o m e n t}{A n g u l a r m o m e n t} = \frac{e}{2 m}$