[PDF] Solution # 07 Solution 07.1: Magnetic field

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Department of Physics

IIT Kanpur, Semester II, 2017-18

PHY103A: Physics IISolution # 07Instructors: AKJ & SCSolution 07.1: Magnetic eld of magnetized cylinder(Griths 3rd ed., Prob. 6.8).FIG. 1:

To nd the magnetic eld, we rst nd the bound currents in the magnetized cylinder (Fig. 1):

GivenM=M^=ks2^impliesJb=rM=1s

@@s (sM)^z= 3ks^z; andKb=M^njR=M^^sjR=kR2^z. Therefore, using Ampere's law, the magnetic eld (Binside) inside the cylinder is found as follows: I B insidedl=0Ienclosed;(1) =)Binside2s=0Z J bda;(2) =)Binside2s=0Z s 0

3ks02s0ds0;(3)

=)Binside=0ks2:(4)

Thus,Binside=0ks2^.

In order to nd the magnetic eld (Boutside) inside the cylinder, we nd the total bound current (there is

no free current anyway):RJbda+RKbdl=RR

03ks2sds+(kR22R) = 0. Consequently, the current enclosed

by an Amperian loop outside the cylinder is zero, implyingBoutside=0.

P.S.: We have neglected the contributions from the surface currents corresponding to the top and the bottom

circular surfaces on the magnetic eld by assuming that the cylinder is practically innitely long. Solution 07.2: Magnetic eld of magnetized looped parallelepiped(Griths 3rd ed., Prob. 6.10).

Please refer to Fig. 2(a). The magnetizationM[(Ms;M;Mz) = (0;M;0)] would generate both volume as well as

1

FIG. 2: Note that we are using overhead arrows for denoting vectorial quantities in these schematic gures in order to avoid

typographical ambiguity. surface bound currents:Jb=rM=1s @@s (sM)^z= (M=s)^zandKb=M^nthat is given by K b=8 >:M ^(^s) = +M^z(for inner surface); M ^(+^z) = +M^s(for upper surface); M ^(+^s) =M^z(for outer surface); M ^(^z) =M^s(for lower surface): Thus, there is a volume current as well as a surface current.

Since the radii of the inner and the outer surfaces are dierent, there is a dierence in the currents on

the inner and the outer surfaces. However, sinceaL, we can ignore the dierence and take the current

to be approximately the same. In the same vein, the current inz-direction due to the volume current is

I b=Ra

0Jb2sds= 2Ra

0 +Ms sds= 2Ma. Obviously,Ibis negligible sinceais small (compare it with the

current inz-direction due to the surface current). These should be kept in mind during further analysis as done below.

Now in order to nd the magnetic eld in the gap, our strategy is to treat the looped parallelepiped as

the superposition of a complete torus plus a square loop with reverse current. Please refer to Fig. 2(b). Had the gap

been not there, the magnetic eld,B, inside the torus can be found using the Ampere's law: I

BdlIenclosed;(5)

=)B2R02RM;(6) =)B0M^:(7)

However, the contribution due to the gap to this eld is missing. The gap can be treated as a square loop with current

in the reverse direction [see Fig. 2(c)]. Since,waL, the gap corresponds to a current loop carrying a current

equal towM. It is well known that the magnetic eld at the center of the square loop carrying currentIis

B=2p20Ia

(^) =2p20wMa ^:(8) 2 Hence, from Eq. (7) and Eq. (8), the net magnetic eld at the center of the gap is B gap=0M^2p20wMa ^;(9) =)Bgap=0M

12p2wa

:(10) 3quotesdbs_dbs19.pdfusesText_25

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