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anonymous
 4 years ago
consider a nonconducting plate of radius 'r' and mass 'm' which has a charge 'q' distributed uniformly over it.the plate is rotated about its axis with an angular speed 'w' ..Show that the magnetic moment 'mu' and angular momentum 'l' of the plate are related as
mu=(q/2m)l
anonymous
 4 years ago
consider a nonconducting plate of radius 'r' and mass 'm' which has a charge 'q' distributed uniformly over it.the plate is rotated about its axis with an angular speed 'w' ..Show that the magnetic moment 'mu' and angular momentum 'l' of the plate are related as mu=(q/2m)l

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2If you just have a circular loop of a circuit with a current of \( I \), then the magnetic moment of that circuit, \( m \) is given by \[ m = IS \] where \( S \) is equal to the vector area of the loop (i.e., a vector with magnitude equal to the area of the loop and in the direction given by the righthand rule wrt the direction of the current. Now use that to write down an integral for the magnetic moment of your set up. Let r be the distance along the radial distance from the center of the disk. Each band between r and r + dr makes up a circuit. This integral looks familiar to the integral to find the moment of a disk. Manipulate it to get both sides of your target equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i did that...but nt getting the eqn...

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2In a band at a distance r from the radius of the disk and of width dr, there is \[ dq = \frac{q}{2\pi r} dr \] of charge. In one time unit, this charge rotates \( \omega \) times. Hence the component of current of that charge is \[ di = \omega dq = \omega \frac{q}{2\pi r} dr \] Write that the disk has radius R. Then the magnetic moment is the integral of all these loops times the area \[ \mu = \int_{disc} (Area) \ di = \int_0^R \pi r^2 . \omega \frac{q}{2\pi r} dr \] \[ = \frac{\omega q }{2} \int_0^R r \ dr \] \[ = \frac{\omega q}{4} R^2 \] Now, the angular momentum of the disk is \[ L = I_{disc} \omega \] where \( I_{disc} \) is the moment of inertia. Write that down and you'll see that you get the expression asked by the question.
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