Magnetic Field and Force (in a solenoid)

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SUMMARY

The discussion focuses on calculating the magnetic field and force exerted on a charged particle within a solenoid. A 200-turn solenoid, 20.0 cm long, carrying a current of 3.25 A, results in a magnetic field of B = 4.08 x 10-3 T. The force on a 15.0 µC charged particle moving at 1050 m/s at an angle of 11.5° is calculated to be F = 1.28 x 10-5 N. The correct formula for the magnetic field inside a solenoid is B = μ0 (n I / l), where n is the number of turns per unit length and l is the length of the solenoid.

PREREQUISITES
  • Understanding of solenoid physics and magnetic fields
  • Familiarity with the formula B = μ0 (n I / l)
  • Knowledge of force calculations involving charged particles
  • Basic algebra for manipulating equations
NEXT STEPS
  • Research the derivation of the magnetic field formula for solenoids
  • Learn about the Lorentz force law and its applications
  • Explore the concept of magnetic flux and its relevance in electromagnetism
  • Investigate the effects of varying current on the magnetic field strength in solenoids
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in electrical engineering or related fields seeking to understand solenoid behavior and magnetic forces on charged particles.

physgrl
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Homework Statement



A 200-turn solenoid is 20.0 cm long and carries a current of 3.25 A. Find the magnetic field. Find the force exerted on a 15.0 µC charged particle moving at 1050 m/s through the interior of the solenoid, at an angle of 11.5° relative to the solenoid’s axis.

Homework Equations



r=mv/qB
torque=IABn (n=turns)
B=(μo/2π)*I/r
F=ILBsin(θ)
F=qvBsin(θ)


The Attempt at a Solution



n=200
l=.2m
I=3.25A

If I want to use torque=IABn to find B I need torque and I do not know how to find it with the length only. The second part I understand once I find the Magnetic Field, but I don't get how to find it with the info given.

the answer key says: B=4.08 x 10-3 T and F=1.28 x 10-5 N
 
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physgrl said:

Homework Statement



A 200-turn solenoid is 20.0 cm long and carries a current of 3.25 A. Find the magnetic field. Find the force exerted on a 15.0 µC charged particle moving at 1050 m/s through the interior of the solenoid, at an angle of 11.5° relative to the solenoid’s axis.

Homework Equations



r=mv/qB
torque=IABn (n=turns)
B=(μo/2π)*I/r
F=ILBsin(θ)
F=qvBsin(θ)


The Attempt at a Solution



n=200
l=.2m
I=3.25A

If I want to use torque=IABn to find B I need torque and I do not know how to find it with the length only. The second part I understand once I find the Magnetic Field, but I don't get how to find it with the info given.

the answer key says: B=4.08 x 10-3 T and F=1.28 x 10-5 N

Hi!

I think you need an extra formula.
The magnetic field inside a long solenoid is given by:
$$B=\mu_0 {n I \over \mathcal{l}}$$
as you can see on wiki: http://en.wikipedia.org/wiki/Solenoid
 
Thanks!
 

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