Finding the Magnetic force on the coil

AI Thread Summary
The discussion focuses on calculating the magnetic force on a coil using basic principles. The user attempts to resolve the force into X and Y components and integrate over a full circle. Feedback suggests verifying the direction of the force and considering the cancellation of horizontal components from diametrically opposite current elements. The user updates their calculations, confirming that the vertical component accounts for multiple turns of the coil. The final expression for the force appears correct, yielding a specific value in the negative Y direction.
Physicslearner500039
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Homework Statement
A voice coil in a loudspeaker has 50 turns of wire and a diameter of 1.56 cm, and the current in the coil is 0.950 A. Assume that the magnetic field at each point of the coil has a constant magnitude of 0.220 T and is directed at an angle of 60.0 Deg outward from the normal to the plane of the coil. Let the axis of the coil be in the y-direction. The current in the coil is in the direction shown (counterclockwise as viewed from a point above the coil on the y-axis). Calculate the magnitude and direction of the net magnetic force on the coil.
Relevant Equations
F=ILBSin(Θ)
1598025773773.png

Surely a tough one, I am doing it from the basics. This is the diagram i tried to draw showing the Force and current I
1598027435817.png


The Length L is the tangent to the circle. The Force F is pointing upwards at ##90 Deg## to the ##\vec B## and also perpendicular to ##\vec L##. I am considering a small length ##\vec dL = r d\theta##. Resolving the force into X and Y components and integrating over the complete ##2\pi##.

##Fx = \int_0^{2\pi} IRB\sin(30) d\theta ##
##Fy = \int_0^{2\pi} IRB\cos(30)d\theta ##

Is my approach correct or completely wrong? Please advise
 
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Check the direction of F in your drawing. The current is going into the page at the point where you draw F.

If you consider the forces on diametrically opposite elements of current, what happens when you add their horizontal components?

Your integral for ##F_y## looks good except for maybe the overall sign. If ##\theta## varies from 0 to ##2 \pi##, how many times do you go around the cylinder?
 
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TSny said:
Check the direction of F in your drawing. The current is going into the page at the point where you draw F.
I am not sure how i have done that mistake, the updated diagram is
1598076875045.png

The horizontal components cancel out. The vertical component is for N turns
##-\int_0^{2\pi} NIRB\cos(30)d\theta##
##-\pi NIRB\sqrt3 = -0.443 \hat j N##
 
Looks right to me.
 
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