A flat, square surface with side length

AI Thread Summary
To calculate the magnetic flux through a square surface in the xy-plane, the magnetic field vector B must be analyzed correctly. The formula for magnetic flux is given by Φ = BAcos(θ), where θ is the angle between the magnetic field and the normal to the surface. The area of the square surface is essential for accurate calculations, as the magnetic field's components must be considered in relation to the surface's orientation. The angle θ is crucial; if the magnetic field is not perpendicular to the surface, the cosine of the angle must be calculated accordingly. Correctly applying these principles will yield the accurate magnetic flux value.
Mdhiggenz
Messages
324
Reaction score
1

Homework Statement



A flat, square surface with side length 3.00cm is in the xy-plane at z=0 .

Calculate the magnitude of the flux through this surface produced by a magnetic field

B=(0.150T)i+(0.350T)j-(0.500T)k

What I'am doing is I know that the magnetic flux= BAcos(theta)

So I multiply my vector B by 0.03m and get (0.0045)i+(0.0105)j, and z=0 so I can ignore that value.

Then they are asking for the magnitude, so √(0.0045)^2+(0.00105)^2= 0.00462Wb , however that is not giving me the correct answer, what am I missing?


Thanks

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
What angle theta are you using for the magnetic flux?

Also, what's the area of a square? Seeing as a Tesla is 1 Wb/m^2, multiplying by 0.03m should get you an answer of Wb/m, not Wb.
 
Theta is not given so I assumed it was just 0degrees
 
Theta is the angle between the "normal" line of your plane and the magnetic field lines. It would be 0 degrees if the magnetic field lines were parallel with the normal (therefore perpendicular), but this is not the case here. You'll have to find the angle from the vector you're given.
 
Le Answer

Use the BAcos(theta) formula to find the flux. But since the Magnetic field vector is given, keep in mind that flux operates perpendicular to the field, so all you really have to do is calculate (B_k)(A)cos(90). And you'll be squaring that number and then taking the root so because you only have one term in the "magnitude finding" process, just drop the negative sign.
 
Lol I finished that class about 8 months ago. Thank you though
 
  • Like
Likes VS9
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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