Torque on a loop within a solenoid?

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roccofitz
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Homework Statement


Question:

A solenoid has of radius 100mm and turns density of 5 turns/mm has a current of 10 Amps flowing in it. There is a loop of radius 15mm and 25 turns within the solenoid. The loop is free to rotate about an axis which is normal to the axis of the solenoid. Determine the torque on the loop when there is 2 Amps flowing in it and its plane is normal to the cross sectional plane of the solenoid.



Homework Equations



magnetic dipole moment mu=NIA
N= number of turns
I=current
A=area

torque=mu X B X=cross product

The Attempt at a Solution



mu =(25)(2)(7.07x10^-4)
mu= 0.03535

I can get the magnetic dipole moment in the loop but after that I am stuck??
 
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That is where I am stuck, I know a formula for B is F=BILsin theta but I don't see how this would work? This is the full question but I've no idea how to get the answer.
 
roccofitz said:
That is where I am stuck, I know a formula for B is F=BILsin theta but I don't see how this would work? This is the full question but I've no idea how to get the answer.
No, that's not the formula for B. You need the formula for the B field inside the solenoid that looks like

B = (A whole bunch of stuff that is given to you by the problem)

Look it up in your textbook or the web.
 
Or better yet, use Ampere's Law to get B yourself. It's great practice, if you don't have much of a handle on using it yet.
 
Ok here's my new attempt right or wrong??

Relevant Equations:

magnetic dipole moment mu=NIA
N=number of turns
I=current
A=area

B=magnetic field= (mu_0)(n)(I)

mu_0=magnetic constant
n=turns density
I=current

Attempt:

mu =(25)(2)(7.07x10^-4)
mu= 0.03535

B=4pi10^-7(5000)(10)=1/50pi

torque=mu X B= muBsin(theta)

I think thetha =90 as the loop is normal to the solenoid axis

therefore torque =muBsin(90)= 0.03535(1/50pi)sin90= 2.22x10^-3 N.m
 
Thank you for all the help :)