A charge Q is uniformly distributed over the surface of...

A charge $\mathrm{Q}$ is uniformly distributed over the surface of non-conducting disc of radius $\mathrm{R}$. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity co. As a result of this rotation a magnetic field of induction $B$ is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure.

Key Concepts

Rotating Charged Systems

A rotating charged object converts static electric charge distributions into moving charges, which in effect behave like currents. This is a fundamental concept in electromagnetism because moving charges generate magnetic fields. The analysis of such systems typically involves finding an equivalent current distribution produced by the rotation of the charge and then using electromagnetic laws to determine the corresponding magnetic field.

Surface Charge Density and Effective Current

When a charge is distributed over a surface, it is characterized by its surface charge density, which is the charge per unit area. In a rotating disc, the surface charge density, together with the angular velocity, determines the effective current that flows in circular paths. This effective current, though distributed over the entire disc, is the source of the magnetic field produced by the rotating charges.

Magnetic Field from Current Loops

A circular current loop produces a well?defined magnetic field at its center, commonly evaluated using the Biot–Savart law. This concept is key because many extended charge distributions can be analyzed by breaking them down into concentric current loops, each contributing to the net magnetic field. Understanding the magnetic field due to a current loop is thus crucial for calculating and predicting the overall field in more complex geometries.

Integration over Continuous Charge Distributions

For objects with a continuous charge distribution, such as the disc, the overall magnetic field is obtained by integrating the contributions from all infinitesimal current elements or rings that make up the object. This integration reflects how the shape and size of the charged object affect the total magnetic field, as changing the geometry alters the contributions from different regions of the object.

Transcript

00:01 Hello stens, in this question we have given that we have a disk of radius r, radius r, okay? and this disk can rotate about its axis with angular velocity omega, okay? and due to this rotation, a magnetic induction, that is, magnetic field is induced at center, whose value is equal to b.

00:20 Now, keeping omega constant, omega constant, and charge, that is q, it is also given, placed on the ring.

00:32 This disk has charge cube okay so omega constant and charge is constant and radius of this disk that is let's say r is variable it means keeping omega constant and charge constant its radius is increasing this radius is increasing now we have to find out variation of magnetic induction at the center due to the condition that is when changing the radius okay so we have given four figures we have to find out which one is correct.

01:05 So first of all whenever this is the ring and magnetic induction at the center of the ring or disk is given by muono type 2r but here r is changing.

01:17 So let's assume a element, thing type element at a distance are at a distance r okay of thickness d r of thickness d r okay now using biodextens servet law we get magnetic induction is equal to mu node upon 2r that is here r equal to small r into i.

01:51 I value can be given as here total charges q and it is distributed our total r radius total r radius so current density will be or chart density will be q upon area that is by r square and we have to determine the current or charge in this elemental ring of at a radius r and thickness is given by let's say dr thickness is dr so in ring element charge will be let's say dq its value will be equal to total charge that is q distributed our area pi r square into area of this ring element its radius is r so this length will be 2 pi r this length will be 2xr into dr this length will be 2 pi r into dr this will be the area this is our length and this is our thickness so total two pi r dr is the area of this element so from viate law mu note upon 2r okay q by r square into total charge that is 2 pi r small r sorry 2 pi small r d r okay because distance of or simply radius of this element is small r and d r is very small okay so this will be 2 pi r d r d r d r d r so this will be 2 pi r d r d r d r integration of this from 0 to capital r at any time t so divide by this t okay or simply this t comes out from because we have given omega that is there will be a time period for one cycle let's say it is equal to t now constant terms can be taken out that is mu node q upon two r square two r square into t into two pi okay this is will become only dr, dr, integration 0 to capital r.

03:53 So this equation becomes our mu node into q upon 2r square into t into 2 pi or dr integration of dr will be r and on putting the limit, this will become 2 pi r.

04:10 Okay, so our this equation 2 and 2 gets cancelled out, 1r, 1 r gets cancelled out...

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