Why Is B°dl Zero at Sections b and d Inside a Solenoid?

AI Thread Summary
The magnetic field contributions at sections b and d inside a solenoid are zero due to the orientation of the magnetic field relative to the differential length element (dl). According to Ampere's Law, the circulation of B°dl must equal zero when dl is perpendicular to the magnetic field. Inside an infinitely long solenoid, the magnetic field is uniform and parallel to the axis, making the contributions from sections b and d null. Therefore, even within the solenoid, the product B°dl equals zero at these sections. This understanding clarifies why the magnetic field's contribution is zero at those points.
member 217843

Homework Statement


Explain why the contribution of B°dl of the magnetic field at sections b and d of the enclosure length must be zero, even inside the solenoid.
cutoutsolenoid.png


Homework Equations


Ampere's Law: ∮ B dl= U_0 I_enclosed N

The Attempt at a Solution


The right and left sides of the loop have a magnetic field that is perpendicular to the solenoid loop. Because of this, it makes no contribution to the circulation B°dl there.
 
Last edited by a moderator:
Physics news on Phys.org
Hi SgtMousse ...

Observe that all the circles in your diagram which reprsent the wires form a solenoid with current in a direction into the plane .Now you must be knowing that the solenoid is infinitely long right? In such a case the components of magnetic field due to the wires which are parallel to the axis of the solenoid add up and all others cancel out...Thus you are left with a uniform (almost uniform if you neglect the edges of the solenoid) magnetic field inside the solenoid parallel to the axis and in this case parallele to the side c... Now in the sections 'b' and 'd' the 'dl' is perpendicular to the magnetic field (which is along the direction of c)... So in those two sides even inside the solenoid B.dl is 0...

Hope you got this clear now...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top