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Magnetic Flux through Solenoid

  • Thread starter tnbstudent
  • Start date
  • #1
13
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Homework Statement


A wire circle of radius 0.050 m is embedded in a solenoid of length 0.20 m with 1000 turns that carries a current of 0.50 A. If a vector that is normal to the plane of the circle makes a 40° angle with the axis of the solenoid, what is the magnetic flux through the loop?



Homework Equations


B (solenoid) = μ*I*n
Flux = B*A*cosθ



The Attempt at a Solution


n=1000/.2
I'm using the loops per unit length
l=.20m
r=0.05m
I=0.5A
I plugged in the values to solve for the magnetic field:
B=(4*∏*10^-7)*(0.5A)*(1000/.2)
B=0.031T

A=(∏r^2)*(l)
A=1.5e-4 m^3

I'm having some trouble visualizing the angle, but I think the angle should be 50?

Can someone point me in the right direction?
 

Answers and Replies

  • #2
168
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Plugged in? You have to do a flux integral.
the flux If the solenoid points alon an axis, z, the angle between the magnetic field and the circle is still 40deg. So
[itex]\Phi = \int \vec{B}_{solenoid}\cdot d\vec{a} = \int |B_{solenoid}|\cos(\theta) da =|B_{solenoid}|\int \cos(\theta)da [/itex]
B goes out in front because it is constant over the circle. So do the surface integral yourself.
I would always suggest to do the solution algebraically first, then insert numbers, and then evaluate your result. Is it realistic and does it diverge at any point.
And always draw your situation, saves a lot of trouble.
 
  • #3
13
0
Thanks - I'm sure you are correct, but we will not get to integrals until a later section. I would like to get the answer using the tools from this section which are the formulas I listed above.
Thanks again - I should have been more specific with my question.
 

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