Recent content by Antoha1

in a triangle,is the angle between the points of contact of the inscribed circle with the sides 120 degrees? From drawings it might be similar to that but I am not sure

Hi, having difficulties in this problem. Looking for ideas. For now, I think the combined voltage of source and coil should be bigger or equal to EMF, for the battery to get charged: ##U+L\frac{dI}{dt}\ge EMF## the current going through the coil, I think, would be...

Hi, I don't have the basics of integral and differential calculus yet, but I dug a little deeper and could this be an expression for the voltage drop between the terminals? ##\varepsilon=\int_{0}^{L}B\omega r\ dr##

I am thinking of using Kirchhoff's laws in two contours. 1st contour would be the small one (E1 and E2) counter clock wise. 2nd contour would go through E3,E2,R (wouldn't go through E1). For 1st contour (CCW): ##E_{1}-E_{2}=r(I_{1}-I_{2})## 2nd contour (CCW): ##E_{3}+E_{2}=I_{3}(r+R)+I_{2}r##...

I do not really know how to solve this problem. For now I just have mathematical expression for ##\varepsilon## EMF. I think this is sources EMF. Correct me if I'm wrong. ##\varepsilon=Blv##. ##l## here is lenght of the rod so ##l=r##. ##v## here is linear speed of the rod...

I have came up with different solution. Could you check it aswell? Do not mind language. I'm adding image.

to calculate momentum (p), do I need to use sum of speed vectors? Maybe someone can help me to solve this problem. For now, my solution looks like this: $$a_{centrifugal}=\frac{v^2}{R}=\frac{F_{Lorenz}}{m}=\frac{qvBsin\alpha}{m}$$ $$\frac{v}{R}=\frac{qBsin\alpha}{m}\implies...

so without absolute of cosB it would be correct?

I think that change in magnetic flux cannot be bigger than this exact ##\phi=BAcos0## initial magnetix flux. Well, imagining from other perspective: If the loop would have been turned not 120 degrees as it's said, but 60 degrees (just to the other side) wouldn't the change in magnetic flux...

from the written formulas, charge is: $$\Delta q=\frac{BA}{R}(cos\alpha-cos(\alpha+\beta))$$ . It is said that in the beginning, the Area was perpendicular to magnetic field lines so the ##\alpha=0## ##\phi_{1}=BAcos0=BA## , Refered to that: $$\Delta q=\frac{BA}{R}(1-cos\beta)$$ but...

Yes. Like frame in DC motor

So, talking about the change in the flux. If the frame in the field is turned 180 degrees, upside down, does the change in the flux become 2*phi(initial

So, talking about the change in the flux. If the frame in the field is turned 180 degrees, upside down, does the change in the flux become 2*phi(initial)?

Hi, school student here. I have a question about magnetic flux. I know it is BAcosa. But can it actually be negative? Based on cosine function it can be, because it is related to it. But based on my logic, magnetic flux is just how much magnetic field (for example lines) goes through the Area...

Hi, I am 10th grade student from Lithuania and I am really interested in physics. Studying further course on my own, but the problem is that I do not have a teacher or helper to ask questions. Hoping to find help here!