Recent content by Siune

Hello, just been reading a little bit about the basics of moving coil meter and one thing is bugging me. First if we have a basic square looped coil in and uniform magnetic field. There is torque ( m x B ) due to the magnetic field and then there is the system/spring attached to the coil...

Homework Statement We have a square loop with side length 2a, at xy plane. Now we know ( I have calculated ) that the magnetic field at the center of the square loop to be H = \frac{2I}{\sqrt{2} \pi a} Now we want to know what is the magnetic field at point P, which is on the axis which...

No problem at all, wouldn't have been the first time I make the mistake on the easiest part of the problem. ^^

Hmm, as the distance vector R from one wire and magnetic field from same wire are perpendicular I get that the x component of magnetic field by that wire so sin(\theta) = \frac{d}{R} and R^2 = x^2 + d^2 ?

Homework Statement We have wires like in the picture and we want to know the value of magnetic field as function of x: NOTE: We want to know the magnetic field at point P ( P is on +x axis ) ( Which is totally arbitary!) The picture is there for just to give the idea. R is the distance from...

Okey, so we can divide the the bottom surface area to n amount of same size segments. Then they all have the same length L. Now as they are parallel and each has resistance R_1 we get R^{-1} = \big( \frac{n}{R_{1}}\big) Now each segment has surface area A_1 =...

Ye, that's why option 1 seemed pretty weird to me. I don't see tho how option 2 is about additivity of resistance. As I just calculated the (bottom) surface area of the hollow sylinder and then proceed to calculate the resistance?

First option: R = 1.7 * 10 ^{-8}\Omega m * \Big(\frac{120 m }{\pi * \big( \frac{0.004 m }{2} \big)^{2}} - \frac{120 m }{\pi * \big( \frac{0.0041 m }{2} \big)^{2}} \Big) = 7,822.. * 10^{-3} \Omega Second option: R = 1.7 * 10 ^{-8}\Omega m * \Big(\frac{120 m }{ \pi \big(...

Homework Statement Hey! We have a copper pipe, which has a outer diameter y = 4.1 mm and inner diameter x = 4.0 mm Length of the pipe is L = 120m and resistivity of copper is ρ = 1.7 * 10^{-8}\Omega Homework Equations Resistance is R = ρ * \frac{L}{A} The Attempt at a...

Yeah, I see it now. :) Thank you so much for your patience!

If the spaceship and pod before the launch had 0 momentum to their mutual centre of mass, the conservation of momentum says: the pod and spaceship must have right after the launch 0 momentum to their mutual centre of mass. So you have 4 variables ( as intial velocities were both 0 ) after the...

Resistance is the reason why conductor/components produces heat/power by equation: P = R I^2, so atleast power/heat the starting motor procudes, goes up. But I don't see that as problem, so hmm.. Current decrease when resistance goes up as the current in circuit is E / R ( of all components...

For motor voltage ( Potential difference ) to increase: 1) Motor resistance can go up ( V = R(motor) * I ), so if R increase -> V increase. but couldn't equally 2) cable resistance to go down? P.S Deeply sorry for my bad understanding about this subject or more especially how to use the...

Okey, so I understand your question so that resistance is ok means that it has the resistance it's given ( for example in this case for motor 0,15 Ω ). So voltage, potential difference across the motor should be with the current I = 66,7 A. V = RI = 0,15 Ω * 66,7 A = 10,0 V and in the...

Homework Statement Hey, I know this is exteremely basic problem and the answer is "easy", but I can't just get the exact idea why it's so. "The starting motor of particular car has a resistance of 0,15 Ω and is powered by 12 V battery through a cable which has a resistance of 30 mΩ ...