A spherical charge distribution of radius, R, contains charge distributed non-un

Question

A spherical charge distribution of radius, R, contains charge distributed non-uniformly as follows: and is spinning with an angular velocity, Show that the infinitesimal current, di, circulating around a circular ring that lies in a plane perpendicular to the z-axis and having coordinates (r. theta) is given by Using the expression for di obtained in a) and modifying the expression for the magnetic field on the axis of symmetry produced by a circular filamentary current that lies in the x-y plane, show that the infinitesimal magnetic field, d B . produced by the current, di. described above at a point, z = z_p. on the z-axis is given by Using the results above and tho charge density, p. given above, obtain an expression for the magnetic field. B, on the axis of rotation (z-axis) at a point. z_0 on the z-axis. Obtain separate expressions for z_0 0. inside and outside of the distribution (-1 different regions). Obtain an expression for the magnetic dipole moment, m. of this spinning charge distribution. For Q = 2 x 10^-5 C. K = 0.02 m. and Omega = 50 rad/s., determine B for points along the z-axis for Zq = 0.02 m. and 0.03 m.

Explanation / Answer

As a charge distribution is spinning, it will naturally induce current.

By definition, current = charge/ time ----(i)

The charge of the ring = ho * d( au)     ----- (ii) , where au is the thickness of the ring

And time = 1/ frequency = 1/ { (2pi )/ (omega)} ------(iii)

Substituting (ii) and (iii) into (i)

So the infinitessimal current di= ( ho) [ (omega ) /(2 pi)] ( d au)   ----- (iv)

Now from the sphere, element d au= r2sin( heta)   . dr. d( heta) . d (phi)

Now for the ring, which lies perpendicular to the z-axis, we need to integarte over the azimuthal coordinate d( phi).

integral of d(phi) from zero(0) to 2(pi), gives 2(pi). ------(v)

Substituting (v) in (iv), we get:

di= ( ho) (omega ) r2 sin ( heta) . dr .( d heta)

Rearranging the terms we get the answer of the first part:

di= ( ho) (omega ) {r sin ( heta) } . { r( d heta) } . dr    -------(vi)

To answer the second part, we need to consider the Biot-Savart law:

dB= [ (mu0) / 4 (pi)] (di) (dl X r) / r3 --------(vii)

The bold forms dB, dl and r are vectors. The magnetic field dB is in z-drection.

And the product of dl with r is also in z-direction.

Substituting the value of di from (vi) in (vii)

dB= [ (mu0) / 4 (pi)] ( ho) (omega ) r2 sin ( heta) . dr .( d heta) (dl X r) / r3 -----(viii)

Now , by the given construction of the ring in this problem, the radial coordinate is r, ( heta) is angle

it makes with the z-direction.

So considering the right -angled triangle formed by the radius of the loop with the point

of observation as the other vertex, we get the perpendicular of the triangle = [z0 - r cos( heta)] ---- (ix)

and the base of the traingle = r sin( heta) ------(x)

Substituing (ix ) and (x) in (viii), we get the desired result for the second part:

dB= [ (mu0) ( ho) (omega ) r4 sin3( heta) . dr .( d heta) ] / [ {r sin( heta)} 2 + {z0 - r cos( heta)} 2] 1/2 z

                        ---------------------(xi)

(c) To get the final expression of the magnetic field, one has to integrate (xi) for the entire element and then

substitute the values of charge distribution. For all cases, ho=0 when r>R

for the other two cases, whether z0>0 or z0 < 0, the term {z0 - r cos( heta)} 2 will have the same value.

This gives the third part of the answer.

The magnetic dipole moment of this spinning charge distribution

= (current I in the loop) X (area of the loop)

From equ (vi),

The current in the loop will be =int [ ( ho) (omega ) {r sin ( heta) } . { r( d heta) } . dr}

And as the charge distribution is spherical, the total area will be the surface area of the sphere= 4pi R3

This gives the answer for the fourth part.

The fifth part is simple but lengthy. So time limit will not permit. So here is a brief description of the process:

One has to substitue the given values of q, omega and other parameters and get the magnitude of the

magnetic field.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.