- #1
tjvogel
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Hey all, I'm new to this site, so bear with me here! First post woot!
A bar magnet with magnetic dipole moment 0.5 A m2 lies on the negative x-axis, as shown in the diagram. A compass is located at the origin. Magnetic North is in the negative z direction. Between the bar magnet and the compass is a coil of wire of radius 1.5 cm, connected to batteries not shown in the picture. The distance from the center of the coil to the center of the compass is 9.9 cm. The distance from the center of the bar magnet to the center of the compass is 22.5 cm. A steady current of 0.836 amperes runs through the coil. Conventional current runs clockwise in the coil when viewed from the location of the compass.
http://tinyurl.com/27o8lfk
Bmagnet= [tex]\mu[/tex]o/4*[tex]\pi[/tex]*(2*[tex]\mu[/tex])/r3
Bcoil= [tex]\mu[/tex]o/4*[tex]\pi[/tex] * (2*pi*R2I)/(z2+ R2)3/2})where mu_o/4 = 1e-7
mu= magnetic dipole moment =.5 A*m2
R= radius= .015
I understand that Bmagnet= negative Bcoil to keep needle pointing north.
I have tried a couple things, but here is what I think it is, and want to make sure that I get it right.
Bmagnet
= 1e-7* [(2*.5)/.225)]
=4.444e-7
Bcoil
=1e-7*(2*pi*.0152I)/(.0152+ .0992)3/2
=1.17727e-7And my assumption that I would then proceed to divide Bmagnet by Bcoil to get the number of loops...
=4.444e-7/1.17727e-7
=3.7752 Turns
Is this right? or am I doing something wrong?
thanks!
(and sorry for the sloppiness/inconsistency of the formulas... I got tired of trying to fix it. still getting used to the formula inputs... esp greek letters)
Homework Statement
A bar magnet with magnetic dipole moment 0.5 A m2 lies on the negative x-axis, as shown in the diagram. A compass is located at the origin. Magnetic North is in the negative z direction. Between the bar magnet and the compass is a coil of wire of radius 1.5 cm, connected to batteries not shown in the picture. The distance from the center of the coil to the center of the compass is 9.9 cm. The distance from the center of the bar magnet to the center of the compass is 22.5 cm. A steady current of 0.836 amperes runs through the coil. Conventional current runs clockwise in the coil when viewed from the location of the compass.
http://tinyurl.com/27o8lfk
Homework Equations
Bmagnet= [tex]\mu[/tex]o/4*[tex]\pi[/tex]*(2*[tex]\mu[/tex])/r3
Bcoil= [tex]\mu[/tex]o/4*[tex]\pi[/tex] * (2*pi*R2I)/(z2+ R2)3/2})where mu_o/4 = 1e-7
mu= magnetic dipole moment =.5 A*m2
R= radius= .015
The Attempt at a Solution
I understand that Bmagnet= negative Bcoil to keep needle pointing north.
I have tried a couple things, but here is what I think it is, and want to make sure that I get it right.
Bmagnet
= 1e-7* [(2*.5)/.225)]
=4.444e-7
Bcoil
=1e-7*(2*pi*.0152I)/(.0152+ .0992)3/2
=1.17727e-7And my assumption that I would then proceed to divide Bmagnet by Bcoil to get the number of loops...
=4.444e-7/1.17727e-7
=3.7752 Turns
Is this right? or am I doing something wrong?
thanks!
(and sorry for the sloppiness/inconsistency of the formulas... I got tired of trying to fix it. still getting used to the formula inputs... esp greek letters)