Calculating current through solenoid?

  • Thread starter Thread starter scoopaloop
  • Start date Start date
  • Tags Tags
    Current Solenoid
AI Thread Summary
To calculate the current in a solenoid using a bar magnet, Faraday's law is applicable, specifically the equation E=N(dphi/dt). Moving the bar magnet towards the solenoid induces a voltage, which can be measured with a voltmeter or multimeter, indicating a change in magnetic flux. The magnetic field of the bar magnet can be expressed as B=(μ₀/4π)(2m/r³), where m is the magnetic dipole moment and r is the distance from the midpoint. The dipole moment m is calculated as m=n(2L)I(πa²), with L being the length from the midpoint to one end and a the radius. Understanding these relationships allows for the determination of current in the solenoid based on the induced voltage.
scoopaloop
Messages
12
Reaction score
0
Not for a class or anything, just out of curiosity, how would you calculate the current in a solenoid, not given a magnetic field, but just using a bar magnet? I know it will just be some form of Faraday's law. E=N(dphi/dt). Maybe not. I'm imagining all I'm really missing is how to calculate the magnetic field of the magnet. Thanks.
 
Physics news on Phys.org
Take the solenoid and connect a voltmeter or multimeter across it. Move the barmagnet towards the solenoid and there should be a deflection in the voltmeter because ##\phi## changes when you keep moving the barmagnet.
or as you know, the bar magnet is also like a solenoid. The expression for magneticfield for barmagnet along axis is $$B=\frac{\mu_0}{4\pi}\frac{2m}{r^3}$$
 
So, just a little clarification, I did a little research online, is M the magnetic strength of each pole and r, I take it is the distance?

Thanks btw.
 
M is the magnetic fipole moment. Here, ##m=n(2L)I(\pi a^2)##
L is the length from midpoint to one end and 'a' is the radius. 'r' is the distance from midpoint to any point along the axis.
 
Awesome. Thank you
 
Susskind (in The Theoretical Minimum, volume 1, pages 203-205) writes the Lagrangian for the magnetic field as ##L=\frac m 2(\dot x^2+\dot y^2 + \dot z^2)+ \frac e c (\dot x A_x +\dot y A_y +\dot z A_z)## and then calculates ##\dot p_x =ma_x + \frac e c \frac d {dt} A_x=ma_x + \frac e c(\frac {\partial A_x} {\partial x}\dot x + \frac {\partial A_x} {\partial y}\dot y + \frac {\partial A_x} {\partial z}\dot z)##. I have problems with the last step. I might have written ##\frac {dA_x} {dt}...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (Second part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8. I want to understand some issues more correctly. It's a little bit difficult to understand now. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. In the page 196, in the first paragraph, the author argues as follows ...
Back
Top