Torque on a loop within a solenoid?

In summary, the magnetic field in the solenoid is 4pi10^-7(5000)(10)=1/50pi. The torque on the loop is 2.22x10^-3 N.m.
  • #1
roccofitz
8
0

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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  • #2
Use the expression for the torque that you posted. What is B inside this solenoid?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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
 
  • #7
That looks about right.
 
  • #8
Thank you for all the help :)
 

Related to Torque on a loop within a solenoid?

1. What is torque on a loop within a solenoid?

The torque on a loop within a solenoid is the rotational force that is applied to the loop due to the interaction between the magnetic field produced by the solenoid and the current flowing through the loop.

2. How is torque on a loop within a solenoid calculated?

The torque on a loop within a solenoid can be calculated using the formula τ = NIABsinθ, where N is the number of turns in the loop, I is the current flowing through the loop, A is the area of the loop, B is the magnetic field strength of the solenoid, and θ is the angle between the direction of the magnetic field and the normal to the loop.

3. What factors affect the torque on a loop within a solenoid?

The torque on a loop within a solenoid is affected by the number of turns in the loop, the current flowing through the loop, the area of the loop, the magnetic field strength of the solenoid, and the angle between the direction of the magnetic field and the normal to the loop.

4. How does the direction of the current affect the torque on a loop within a solenoid?

The direction of the current flowing through the loop will determine the direction of the torque. If the current is flowing in the same direction as the magnetic field, the torque will be in one direction. If the current is flowing in the opposite direction, the torque will be in the opposite direction.

5. What is the practical application of torque on a loop within a solenoid?

The torque on a loop within a solenoid has many practical applications, such as in electric motors, generators, and various types of sensors. It is also used in electromechanical devices for controlling and manipulating objects, such as in robotics and industrial machinery.

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