A thin, infinitely long cylindrical conducting shell of radius R 1 = 5 cm carrie

Question

A thin, infinitely long cylindrical conducting shell of radiusR1 = 5 cm carries a currentI1 = 18 mA in the +z-direction (out ofthe page). This current is uniformly distributed in the azimuthaldirection about the z-axis. A second thin, infinitecylindrical shell of radius R2 = 10 cm carriesa current I2 = 31 mA, also in the+z-direction, also uniformly distributed in the azimuthaldirection.

(Note: For purposes of illustration, thethicknesses of the two cylindrical shells have been exaggerated inthe above figure. By "a thin shell", one means that the shellthickness is negligible compared to its mean radius, and thereforethe inner and outer radii of the shell can be consideredequal.)

(a) What is the magnitude of the net magnetic field at adistance r = 2.5 cm from the axis of symmetry?

B = T     *
0   OK

(b) What is the magnitude of the magnetic field at a distancer = 20 cm from the axis of symmetry?

B = T     *
4.9E-8   OK

HELP:  Again, as in part (a), you canuse symmetry and Ampere's law to find the field at this radius,which is outside of both cylindrical shell conductors, using theRight Hand Rule (RHR) to get the algebraic signs of the currentsI1 and I2 correct.
HELP:  Equivalently, you can use theresults of Ampere's law for the magnetic field outside asingle infinite cylindrically symmetric shell of current (same asthat of an infinite wire along the z-axis) andsuperposition to find the net field at this radius.

(c) What is the magnitude of the magnetic field due to thecurrent I1 alone at r =R2?

B = T     *
3.6E-8   OK

HELP:  Use Ampere's law for currentI1 alone with a circle of radius r =R2 about the z-axis as the path.
HELP:  In this calculation you ignorecompletely the presence of current I2 since youare interested in the field caused by I1 alone.The presence of the second conductor of current makes nodifference!

(d) Calculate the magnitude of the magnetic force per unitarea on the outer shell due to currentI1.

dF/dA = N/m2    
-6.2E-8   NO

HELP:  To calculate the force perunit area, you need an expression for a small area on the side ofthe cylinder having radius R2. In writing this,you want to use coordinates that respect the cylindrical symmetryof the problem. Namely, take a segment of the cylinder of area oflength Dz along the z-axis and "width"R2Dq, where Dq is the anglebetween two rays drawn perpendicularly from the z-axis tocylinder 2. Thus, DA =Dz R2Dq.(If you prefer the more elegant notation of infinitesimals, this iswritten d2A = dz R2dq.)
HELP:  Calculate the force on thecurrent in this area of cylinder 2, using the standard expressionfor a straight line segment, using the fact that the amount ofcurrent carried by this area is I2(Dq/2p).Divide your answer for the force by the area to get the force perunit area. The vector cross product that you have taken in applyingthe force law gives you the direction. This will be useful in thenext part.

(e) Calculate the magnitude of the net magnetic force perunit length in the z-direction on the outer shell dueto current I1.

dF/dz = N/m     *
0   OK

A thin, infinitely long cylindrical conducting shell of radiusR1 = 5 cm carries a currentI1 = 18 mA in the +z-direction (out ofthe page). This current is uniformly distributed in the azimuthaldirection about the z-axis. A second thin, infinitecylindrical shell of radius R2 = 10 cm carriesa current I2 = 31 mA, also in the+z-direction, also uniformly distributed in the azimuthaldirection. (Note: For purposes of illustration, thethicknesses of the two cylindrical shells have been exaggerated inthe above figure. By a thin shell, one means that the shellthickness is negligible compared to its mean radius, and thereforethe inner and outer radii of the shell can be consideredequal.) (a) What is the magnitude of the net magnetic field at adistance r = 2.5 cm from the axis of symmetry? B =0 T * 0 OK (b) What is the magnitude of the magnetic field at a distancer = 20 cm from the axis of symmetry? B =4.9E-8 T * 4.9E-8 OK HELP: Again, as in part (a), you canuse symmetry and Ampere's law to find the field at this radius,which is outside of both cylindrical shell conductors, using theRight Hand Rule (RHR) to get the algebraic signs of the currentsI1 and I2 correct. HELP: Equivalently, you can use theresults of Ampere's law for the magnetic field outside asingle infinite cylindrically symmetric shell of current (same asthat of an infinite wire along the z-axis) andsuperposition to find the net field at this radius. (c) What is the magnitude of the magnetic field due to thecurrent I1 alone at r =R2? B =3.6E-8 T * 3.6E-8 OK HELP: Use Ampere's law for currentI1 alone with a circle of radius r =R2 about the z-axis as the path. HELP: In this calculation you ignorecompletely the presence of current I2 since youare interested in the field caused by I1 alone.The presence of the second conductor of current makes nodifference! (d) Calculate the magnitude of the magnetic force per unitarea on the outer shell due to currentI1. dF/dA =-6.2E-8 N/m2 -6.2E-8 NO HELP: To calculate the force perunit area, you need an expression for a small area on the side ofthe cylinder having radius R2. In writing this,you want to use coordinates that respect the cylindrical symmetryof the problem. Namely, take a segment of the cylinder of area oflength Dz along the z-axis and widthR2 Dq, where Dq is the anglebetween two rays drawn perpendicularly from the z-axis tocylinder 2. Thus, DA =Dz R2Dq.(If you prefer the more elegant notation of infinitesimals, this iswritten d2A = dz R2dq.) HELP: Calculate the force on thecurrent in this area of cylinder 2, using the standard expressionfor a straight line segment, using the fact that the amount ofcurrent carried by this area is I2(Dq/2p).Divide your answer for the force by the area to get the force perunit area. The vector cross product that you have taken in applyingthe force law gives you the direction. This will be useful in thenext part. (e) Calculate the magnitude of the net magnetic force perunit length in the z-direction on the outer shell dueto current I1. dF/dz =0 N/m * 0 OK

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