Recent content by palaphys

also, I think I have found the answer to my question. let ##B_2## be the magnetic field due to the rest of the cylinder and ##B_1## be that due to the current element alone. then, if we assume that the magnetic field magnitude does NOT change considerably inside and outside the cylinder, then...

https://en.wikipedia.org/wiki/Magnetic_pressure

Here is my solution: Consider a thin cylindrical shell of radius ##R## carrying total current ##I##, cut lengthwise into two identical semicylindrical halves of length ##l##. We can treat the flat cut face between the halves as a surface experiencing magnetic pressure ## p=\frac{B^2}{2\mu_0}, ##...

Agreed. This is simpler. Thanks

I think I will do it this way- for the entire wire covering the arc, the lengthwise turn density is N/(arc length). here arc length will be ## s= R\phi_{\max} = R\pi/3 ## based on the geometry of the problem. so ##dN/ds= N/s= \frac{3N}{\pi R}## where ##ds=R(d\theta) ## where theta is the polar...

I'm not able to edit it right now, but thanks for pointing it out. how is that possible? even their solution has included the effect of gravity right

I have tried to proceed with two methods, which agree with each other i) Energy analysis If we use work energy theorem, ## \sum W = \Delta K ## So, ## W_{\text{bat}} + W_{\text{field}} + W_{\text{grav}} = K_f \quad \text{(at } x = d/2 \text{)} ## Battery work: ## W_{\text{bat}} = V \cdot (C_i...

yes It must be proportional to some quantity. I'm trying to consider its relation with an angle theta, i.e polar angle

this is the figure provided. My approach was to consider the field due to an elemental ring-shape conductor, using the standard formula ##dB = \frac{\mu_0 i r^2 \ dN}{2(x^2 + r^2)^{3/2}} ## where r is the radius of the elemental ring, x is the distance from the center of the hemisphere to that...

I think I am satisfied with my solution, so I will continue using this kind of approach for such questions..

I agree that state 1 is the intended state. Also when I thought a bit more about this question, I think I understood it now, they are just saying that the energy of the bound K shell can be thought of as -27.5KeV in a way, so we can use ## E_L- E_K = E_{K_{alpha}} ## hence the "energy of the...

The question seems very simple, and its solution is also very simple. But what bothers me is the wording- "energy of Molybdenum atoms" what does that mean?? energy relative to what? also how can the energy of a bound system be positive? and I have never seen "energy of atoms" being used as a...

This post makes everything clear. Very high quality. All my doubts on this question got cleared! for a) I will take the direction of normal as rightward (as shown in diagram) b) ##\phi_{+q} = \frac{q}{10\varepsilon_0}## and ## \phi_{-q} = \frac{2q}{10\varepsilon_0}## c)...

1.##\frac{q}{\epsilon_0}## 2. ##\frac{-q}{\epsilon_0}## 3a)0 b)0

yes I wanted to do this