Recent content by Tomi Kolawole

Which direction is the B pointing in? i assume you use right hand rule for this? your solution makes a lot of sense now

I think because the angle at which The charged particle will be at a certain time changes because the particle is constantly moving upwards with velocity v*. so wouldn't it be unwise to assume the angle between them is 90 if its constantly changing sir?

that may be the case but since i converted my sin(theta) to its compenent opp/hypotenus which is R/r =R/(D^2+R^2) wouldn't that account for whatever theta might be? even if its 90 degrees? i only converted my variables to make integration easier because r changes as the particle moves.

They didnt say which is greater than the other in the question stated above, so how can i know?

my cross product was (dsxr) and because its bcross product the solution is ds*rsin(theta). and then i converted all my variables into units of length through pythagoran theorem.So r^2 became (D^2+R^2) and sin(thetha) became R/(d^2+r^2)^(1/2) that's how i evaluated my cross product? Can you tell...

if i integrate ds, which is each infinitesmal length of the wire, i should get the circumference no? which is 2*PI*R the length of the wire

The formula of bio savart has (i*ds x r)/r^3 . <<<<<That is the DS i am referring to sir

yes sir that's what i meant isn't that the length of the wire? so will that be 2*pi*R?

i solved for the magnetic field and got (u*i*DS*R)/(4*pi*(R^2+D^2)^(3/2)).Is the DS=2piR? if it is my problem is solved but i don't know why it would be 2*PI*R. with R being the radius of my circular wire.

Homework Statement Homework Equations [/B]The Attempt at a Solution I soled for the integral of the magnetic field but i don't know what bounds to intergrate over and also i what is DS in this case? its a point charge so shouldn't bio savart have q(v*b) instead of i(ds*r)?[/B]

Homework Statement Homework EquationsThe Attempt at a Solution I have already solved for the magnetic fields everywhere AND THE magnetic field at R>F = (U/2*PI*R)*(I2-I1) My only problem is i know i use the right hand rule to determine the direction of the magnetic field B as clockwise or...

also is the charge density of the entire sphere the same at any specific point of the same sphere? what conditions would make this true?

i figured it out and i got the right answer.except i have a -q instead of just a +q in the final answer equation.Is this because its an insulator so the sign of the charge doesn't matter?

why is it that there is no charge inside the sphere of radius A? because to calculate the electric field from the centre to My gausian saurface between A AND 3A i will have to neglect the distance r<A during my integration.

outside surface