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Calculate the magnetic force of attraction between the northern and southern hemispheres of a spinning charged spherical shell.

Short Answer

Expert verified

The magnetic force of attraction is 4022R4.

Step by step solution

01

Significance of the magnetic force

The magnetic force mainly arises amongst the particles having electrically charged and in also motion. However, the magnetic force is mainly the repulsion or attraction amongst the charged particles.

02

Determination of the magnetic attraction force

The equation of the surface element鈥檚 force is expressed as:

B=12(Bin+Bout)

Here, is the surface element鈥檚 force, Binand Boutare force inside and outside of the spinning shell.

The equation of the magnetic scalar potential鈥檚 curl of the inside of the shell is expressed as:

Aout=0R43sinr2^

Here, Ainis the magnetic scalar potential鈥檚 curl of the inside of the shell, Ris the radius, 0is the permeability constant, is the angular velocity, is the variability measurement, is the distance from the field, is the angle subtended and localid="1657519378591" ^is the cap of the angle.

The cross product of with the Ainis expressed as:

Ain=Bin=1rsin0R3rsin2r^-1rr0R3rsin2^

The cross product of with the Aoutis expressed as:

Aout=Bout=1rsin0R43r2sin2r^-1rr0R43r2sin2^=230R41r3(2cosr^+sin^)

The equation of the average surface of the sphere is expressed as:

Bavg=12(Bin+Bout)r-R

The above equation can be reduced as:

Bavg=12130R41R3(2cosr^+sin^)+230R(cosr^-sin^=16R(4cosr^-sin^)

The equation of the force on the differential surface element is expressed as:

dF=dqvBavg=dSvBavg 鈥(颈)

Here, dqv is the differential surface element.

The equation of the velocity of the shell is expressed as:

v=Rsin^

The differentiation of the above equation is expressed as:

dS=R2sindd

Substituting Rsin^for vand R2sinddfor dSin the equation (i).

dF=16022R4sin2dd^(4cosr^-sin^)=16022R4sin2(4cos^-sinr^)dd

The equation of the z component of the force of the shell is expressed as:

dFz=dFz^

鈥(颈颈)

Here, dFis the differential force and z^is the position vector.

The equation of the angle subtended by the force is expressed as:

^=coss^-sinz^

The equation of the position vector in the zdirection is expressed as:

localid="1658556502484" r^=sins^+cosz^

Substituting the values in the equation (ii).

dF=16022R4sin2(4cossin-cossin)d=12022R4sin3cosdd

Double integrating the above equation with the integral limits can be expressed as:

F=12022R4020/2sin3cosdd=022R40/2sin3胃肠辞蝉诲胃

鈥(颈颈颈)

The change in the variables can be expressed as:

x=sindx=cosd

Substitute the value of the integral in the equation (iii).

F=022R401x3dx=022R4x44=022R414=4022R4

Thus, the magnetic force of attraction is 4022R4.

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Most popular questions from this chapter

A current flows to the right through a rectangular bar of conducting material, in the presence of a uniform magnetic fieldBpointing out of the page (Fig. 5.56).

(a) If the moving charges are positive, in which direction are they deflected by the magnetic field? This deflection results in an accumulation of charge on the upper and lower surfaces of the bar, which in turn produces an electric force to counteract the magnetic one. Equilibrium occurs when the two exactly cancel. (This phenomenon is known as the Hall effect.)

(b) Find the resulting potential difference (the Hall voltage) between the top and bottom of the bar, in terms ofB,v(the speed of the charges), and the relevant dimensions of the bar.23

(c) How would your analysis change if the moving charges were negative? [The Hall effect is the classic way of determining the sign of the mobile charge carriers in a material.]

A circular loop of wire, with radius , R lies in the xy plane (centered at the origin) and carries a current running counterclockwise as viewed from the positive z axis.

(a) What is its magnetic dipole moment?

(b) What is the (approximate) magnetic field at points far from the origin?

(c) Show that, for points on the z axis, your answer is consistent with the exact field (Ex. 5.6), when z>>R.

(a) By whatever means you can think of (short of looking it up), find the vector potential a distance from an infinite straight wire carrying a current . Check that .A=0and A=B.

(b) Find the magnetic potential inside the wire, if it has radius R and the current is uniformly distributed.

(a) Prove that the average magnetic field, over a sphere of radius R,due to steadycurrents inside the sphere, is

Bave=042mR3

wheremis the total dipole moment of the sphere. Contrast the electrostatic

result, Eq. 3.105. [This is tough, so I'll give you a start:

Bave=143R3Bd

WriteBasA ,and apply Prob. 1.61(b). Now put in Eq. 5.65, and do the

surface integral first, showing that

1rda=43r'

(b) Show that the average magnetic field due to steady currents outsidethe sphere

is the same as the field they produce at the center.

Question: Using Eq. 5.88, calculate the average magnetic field of a dipole over

a sphere of radius Rcentered at the origin. Do the angular integrals first. Compare your answer with the general theorem in Prob. 5.59. Explain the discrepancy, and indicate how Eq. 5.89 can be corrected to resolve the ambiguity at . (If you get stuck, refer to Prob. 3.48.) Evidently the truefield of a magnetic dipole is

Bdip(r)=04蟺谤3[3(mr^)r^-m]+203m3(r)Bdip(r)=04r3[3mr^r^-m]+203m3(r)

Compare the electrostatic analog, Eq. 3.106.

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