Finding the Magnetic Field of a Point Charge Using Biot-Savart Law

[Tomi Kolawole]

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]

 

[vela]

The Attempt at a Solution

I solved 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]

It's not clear what you've done. You say you solved the integral of the magnetic field, but you later say you don't know what goes into it. Please show your actual work instead of just describing it.

Can you explain why the point charge experiences a force? And what is the Biot-Savart law used for here? If you can answer those questions, you should be able to figure out the answer to your last question. And you really need to know the answer to those questions to solve the problem.

 

[Tomi Kolawole]

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.

 

[vela]

$d\vec{s}$ is an infinitesimal quantity. Is that what you mean by DS?

 

[Tomi Kolawole]

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

 

[Tomi Kolawole]

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

 

[vela]

The wire isn't infinitesimal, so no, $d\vec{s}$ isn't the length of the wire. If you were to integrate $ds = \| d\vec{s} \|$, then you'd get $s$, which would be the length of the wire. Note that $d\vec{s}$, $ds$, and $s$ aren't interchangeable. They mean different things.

Back up for a second. Remember the integrand of the Biot-Savart law is a vector, so you have to add vector-wise all the infinitesimal contributions to the magnetic field at the location of the charge. What happens when you do that and integrate around the entire ring? What components will be in the final sum?

 

[Tomi Kolawole]

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

 

[vela]

Yes, that's what I said in my previous post. But integrating $d\vec{s}$ and integrating $ds$ isn't the same thing. One is a vector while the other is the length of the vector. Integrating $d\vec{s}$ around the ring would give you 0, the displacement between the starting and ending points, but integrating $ds$ would give you $2\pi R$, the distance around the path.

Anyway, I'm asking you to back up because your integrand is incorrect. You can't simply ignore the cross product in the integrand.

 

[Tomi Kolawole]

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 me what was wrong with my approach sir

 

[vela]

$\theta$ is the angle between $d\vec{s}$ and $\vec{r}$, the vector connecting the infinitesimal piece of the ring to the point where the charge is, right? Isn't the angle between them $90^\circ$?

 

[Tomi Kolawole]

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.

 

[Tomi Kolawole]

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?

 

[vela]

I was mistaken. Your integrand is correct, but you apparently stumbled onto the right one accidentally because your reasoning to get it is incorrect. $\sin 90^\circ=1$ and is clearly not equal to $R/r$.

There are two angles in this problem (in calculating the magnetic field at the location of the charge). One is the angle $d\vec{B}$ makes with the the horizontal axis. Let's call that $\phi$. And then there's the angle between $d\vec{s}$ and $\vec{r}$, which you called $\theta$ and which is $90^\circ$. So $\| d\vec{s} \times \vec{r} \| = \| d\vec{s} \| \| \vec{r} \| \sin\theta = ds\,r$.

Now when you integrate $d\vec{B}$, only the horizontal component, $dB_z = dB \cos\phi$, survives. You should be able to explain why this is the case. And $\cos\phi = R/\sqrt{R^2+D^2}$, which, again, you should be able to convince yourself is correct.

 

[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

 

[vela]

Yeah, use the right-hand rule to determine the direction of $d\vec{B}$. It's going to rotate as you go around the ring.