- #1
maximus123
- 50
- 0
Hello everyone, this is probably a simpler problem than I think but here it is
My argument is that the length of the solenoid is the number of turns times the diameter of the wire so
[itex]
B=\mu_{0}\times n\times I\\\\
=\mu_{0}\times \frac{N}{L}\times I\\\\
=\mu_{0}\times \frac{N}{N\times d}\times I\\\\
=\mu_{0}\times \frac{1}{d}\times I\\\\
B=4\pi\times 10^{-7}\times\frac{1}{300\times 10^{-6}}\times 10 = 0.042\textrm{ T}
[/itex]
So one layer gives 0.042 Tesla so to get 10 Tesla you'd need [itex]10/0.042\approx 238[/itex] layers.
I don't have a great deal of confidence in this argument but I'm struggling to see how else to approach this question. Any help would be greatly appreciated. Thanks.
You are designing an electromagnet capable of producing a magnetic induction field of 1 T
by winding a solenoid around a long cylinder. A solenoid is typically wound by starting
at one end of the cylinder and placing loop after loop directly next to each other. Once
the other end of the cylinder is reached the first winding layer of the solenoid is complete
and the next layer is started by again placing loop after loop next to each other. You are
using 300 [itex]\mu[/itex]m diameter wire and the current needed to reach 1 T should not exceed 10 A.
What is the minimum number of winding layers required?
My argument is that the length of the solenoid is the number of turns times the diameter of the wire so
[itex]
B=\mu_{0}\times n\times I\\\\
=\mu_{0}\times \frac{N}{L}\times I\\\\
=\mu_{0}\times \frac{N}{N\times d}\times I\\\\
=\mu_{0}\times \frac{1}{d}\times I\\\\
B=4\pi\times 10^{-7}\times\frac{1}{300\times 10^{-6}}\times 10 = 0.042\textrm{ T}
[/itex]
I don't have a great deal of confidence in this argument but I'm struggling to see how else to approach this question. Any help would be greatly appreciated. Thanks.