Recent content by Zatman

Yes, it is only valid for a constant force? Which in this question is not actually the case, I notice.

P = \frac{dW}{dt} = \frac{d(Fx)}{dt} = F\frac{dx}{dt} = Fv P = F_{net}v = \frac{\mu_0IvabI'}{2\pi D(D+b)} Now I think I can substitute the expression for I': I' = \frac{emf}{R} = \frac{\mu_0Ivab}{2\pi RD(D+b)} Which gives the correct answer. I see my mistake now; I somehow managed to...

Okay, so the only things that will be different are (1) the sign, since the current is in the opposite direction. (2) D --> D+b So the force on the far side of the loop is F_2 = -\frac{\mu_0aII'}{2\pi (D+b)} The net force is then the sum of the two forces: F_{net} =...

Hi TSny, thanks for your reply. I have: F = \displaystyle\int^a_0 I'\frac{\mu_0I}{2\pi D}\ dl F = \frac{\mu_0aII'}{2\pi D}

Apologies for resurrecting the thread, but I have the exact same question set so it seemed pointless to make a new topic. I have done the first part (calculating the emf), but do not know how to find the force. I have tried using F = I dl x B with I = emf/R. F = \frac{\mu_0Iva}{2\pi...

Oh, I had theta and phi mixed up. So that gives ∫∫r.ndS = ∫(2a3)dø from 0 to 2π, hence = 4πa3

Good point, thank you. But then I get ∫∫a3sinθdθdø Which evaluates to zero. Are the limits incorrect?

Homework Statement Evaluate ∫∫r.ndS where r=(x,y,z) and n is a normal unit vector to the surface S, which is a sphere of radius a centred on the origin. 2. The attempt at a solution I decided to use polar coordinates. The radius of the sphere is clearly constant, a. So a surface...

Well, don't I feel silly now. So including the radial velocity gives a term with dr/dt. I set the energy equation equal to the initial energy at t=0 {i.e. (1/2)mv02 - GMm/R}. Then at the apogee, r=2.5R and dr/dt = 0, solve for v_0 which gives the required result. Thank you for pointing out my...

Homework Statement It is required to put a satellite of mass m into orbit with apogee 2.5 times the radius of the planet of mass M. The satellite is to be launched from the surface with speed v0 at an angle of 30° to the local vertical. Use conservation of energy and angular momentum to...

Yes, that was conservation of angular momentum, so that k=\frac{L}{m}

Oh, okay, I've got it now (multiply by \frac{\dot{r}}{r}). Thank you so much for your help, voko.

So I actually get: \ddot{r} - r\dot{\theta}^2 = -\ddot{r} - g 2\ddot{r} - r\dot{\theta}^2 + g = 0 Now substituting \dot{\theta}, from above post: 2\ddot{r} - \frac{k^2}{r^3} + g = 0 2r\ddot{r} - \frac{k^2}{r^2} + gr = 0 So still not correct.

...exact. So this gives: r^2\dot{\theta} = k where k is a constant. Rearranging for \dot{\theta} and substituting into the first equation: \ddot{r} - r\frac{k^2}{r^4} = \ddot{r} - g where I have also used the other eqn. to eliminate T. EDIT: just realized this part is wrong, now...

There was actually a previous part of the question asking for the derivation of the same equation using conservation laws, so I can't use that here. I don't understand how that equation can be integrated.