A long hollow cylindrical conductor (inner radius = 3.0 mm, outer radius = 5.0 m

Question

A long hollow cylindrical conductor (inner radius = 3.0 mm, outer radius = 5.0 mm) carries a current of 12 A (into the page) distributed uniformly across its cross section. A long wire which is coaxial with the cylinder carries a current of 16 A in the opposite direction (out of page). To find the magnitude of the magnetic field 4.0 mm from the axis, follow the following steps and show your work. What is the current density J in the hollow conductor? 8a._______Draw an appropriate amperian loop and choose a direction of integration. Apply Ampere's Law and calculate the magnetic field (magnitude and direction) at a point 4.0 mm from the axis.

Explanation / Answer

Inside the solid shell, current density is constant.

Current density, J = I/A = I/(*(b2-a2))

According to Ampere's law,

B*2**r = 0*Ienc

where Ienc is the current enclosed in the area upto radius r.

Ienc = J**(r2-a2)

B = 0*I*(r2-a2)/(2**r*(b2-a2)) = 4**10-7*12*(16-9)/(2**4*10-3*(25-9)) = 2.63*10-4 T

This is the field due to current in the hollow cylinder.

Magnetic field due to the wire = 0*I/(2**r) = 4**10-7*12/(2**4*10-3) = 6.0*10-4 T

They are both in opposite directions.

Net magnetic field = 6.0*10-4 T - 2.63*10-4 T = 3.37 *10-4 T

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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