A 3000-turn solenoid has a length of 60 cm and a diameter of 8 cm. If this solen

Question

A 3000-turn solenoid has a length of 60 cm and a diameter of 8 cm. If this solenoid carries a current of 5.0 A, find the magnitude of the magnetic field inside the solenoid by constructing an Amperian loop and applying Ampere's Law. How does this compare to the magnetic field of the earth (0.5 gauss). [Ans: 0.0314 T. or 314 gauss, or about 600 times the magnetic field of the earth]. We make a magnetic field in the following way: We have a long cylindrical shell of nonconducting material which carries a surface charge fixed in place (glued down) of sigma C/m^2, as shown in Figure 9.13.9 The cylinder is suspended in a manner such that it is free to revolve about its axis, without friction. Initially it is at rest. We come along and spin it up until the speed of the surface of the cylinder is v_0. What is the surface current K on the walls of the cylinder, in A/m? [Ans: K = sigmav_0.] What is magnetic field inside the cylinder? [Ans. B =Mu_0K = Mu_0sigmav_0, oriented along axis right-handed with respect to spin.] What is the magnetic field outside of the cylinder? Assume that the cylinder is infinitely long. [Ans: 0].

Explanation / Answer

part a:

magnetic field inside a solenoid is given by

B=mu*(N/L)*I

where mu=magnetic permeability of the medium (in this case , air)=4*pi*10^(-7)

N=number of turns=3000

L=length of the solenoid=0.6 m

I=current =5 A

then magnetic field=0.0314 T=314 gauss

hence it is 314/0.5=628 times magnetic field of the eart

part b:

current=charge/time

surface current=current per unit length

=charge/(time*length)

=(charge/length^2)*(length/time)

=charge density*linear speed

=sigma*v0

part c:

magnetic field =mu*magnetic field intensity

as per ampere's law,

line integral of magnetic field intensity=current enclosed

==>magnetic field intensity=current per unit length
=surface current density

hence magnetic field=mu*surface current desnity=mu*sigma*v0

part d:

as the cylinder is infinitely long, the return path for the current flowing through the cylinder exist along the cylinder itself.

so, if you consider any ampere path containing the cylinder,

total current enclosed=0

hence outside the cylinder, as per ampere's law, magnetic field intensity=0

==>magnetic field density=mu*magnetic field intensity=0

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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