Question

Problem involving magnetism and current! Any help would be greatly appreciated! :)

Answer 1

Many living things can sense magnetic fields and have an innate tendency to face magnetic north regardless of wind strength, time of day, or position of the sun. In an experiment, a gold ring ID collar of radius 0.06m was attached on a fox. It takes the fox 0.2s to move from a region in which the collar is perpendicular to earth's magnetic field of 5.0E-5 T to one in which the collar is parallel to Earth's magnetic field.. What current, if any, is induced in the collar?

Answer 2

@Michele_Laino @IrishBoy123

Answer 3

Use Faraday's law of induction.

Answer 4

@Vincent-Lyon.Fr could you elaborate on that please? I'm not saying give me the answer, but I am completely stuck on how to approach this problem =/

Answer 5

the change of cocatenated magnetic flux with the loop is:
\[\Delta \Phi = \pi {r^2}B\]
where \(r\) is the radius of the loop, and \(B\) is the magnetic field.
The time in which such magentic flux change occurred is \( \Delta t=0.2\) seconds. So the electrical current \(I\) of the loop, has the subsequent magnitude:
\[I = \frac{1}{R}\frac{{\Delta \Phi }}{{\Delta t}} = \frac{{\pi {r^2}B}}{{R\Delta t}}\]
where \(R\) is the electrical resistance of the loop or collar around the fox

Answer 6

In order to find \(R\) we can apply this formula:
\[R=\rho \frac{{2\pi r}}{s}\]
where \(\rho\) is the specific resistance of gold, which is tabulated, and \(s\) is the croos sectional area of the collar

Answer 7

cross* sectional area

Answer 8

nevertheless, from your data, we aren't to find the value of \(R\) since, we don't know the value of \(s\)

Answer 9

oops..we aren't able to find...

Answer 10

Hmm, are you sure that there isn't another way to calculate the current? It seems as if the question is pushing us to find a numerical answer rather than a function. But perhaps not?

Answer 11

if we apply the Faraday's law, we are able to find the voltage along the collar, nevertheless, if we want to compute the magnitude of the corresponding current, we need to know the electrical resistance of the collar

Answer 12

Hmm, what if we were able to consider it as a very thin wire?

Answer 13

I think that a collar has to have a cross sectional area

Answer 14

oops.. another typo: concatenated*

Answer 15

Hmm, okay. Maybe it's best to go along the route with Faraday's law and voltage/emf?

I know that \[emf=- \frac{ d\Phi_M }{ dt }\]and \[\Phi_M=\int\limits B*dA\]

Answer 16

what is the numerical result, please?

Answer 17

I don't have the answer =/

Answer 18

yes! your procedure above is correct, nevertheless, we can apply the Faraday's law in integral form

Answer 19

How do you mean integral form?

Answer 20

I mean without using differentials

Answer 21

like this:
\[E = \frac{{\Delta \Phi }}{{\Delta t}}\]

Answer 22

Ah okay. So once I have the emf, then
\[V_{emf}=IR\] right? But we're not given R.. hmm

Answer 23

yes! Your point of view is right!

Answer 24

Hmm, if I say \[emf=vBd\]then I could relate the two as\[vBd=-\frac{ B (\Delta A) }{ \Delta t }\]Right?
Looks like the B cancels out anyway. But then I don't know v so it looks like I'm still stuck anyway..

Answer 25

please the left side is the e.m.f. at the ends of a conductor which is moving inside a magnetic field. Whereas your problem can be modeled by this drawing:

Answer 26

Hmm, okay, where can I go after this?

Answer 27

I see that \(\Phi_M\) would be 0 after, since A is parallel to B, correct?

Answer 28

yes! Before the flux is \(\Phi = \pi {r^2}B\), and after the flux is \(\Phi =0\), so its change is:
\[\Delta \Phi = 0 - \pi {r^2}B = - \pi {r^2}B\]

Answer 29

of course, with the word "flux", I mean the concatenated flux with the loop

Answer 30

Right, which means that:
\[emf=\frac{ \pi r^2B }{ \Delta t }\]

Answer 31

correct!

Answer 32

Okay, so once I have the emf, how can I relate this back to current?

Answer 33

we can apply the Ohms's law

Answer 34

But we don't know R? >_<

Answer 35

Oh, we would just look that up right?

Answer 36

Yes, I know! Nevertheless, the collar is made by gold, which is a metal, more precisely a noble metal, and the conduction inside metals is modeled by the Ohm's law

Answer 37

I found that \(\rho_{Au}=22.1 \ n\Omega \ m\)

So then we can just plug everything in at this point, right?\[\frac{ \pi r^2 B }{ (22.1\times 10^{-9})t }=I\]

Answer 38

@Michele_Laino I got 127.94 amps -- does this seem reasonable?

Answer 39

I think that it is too high

Answer 40

I thought so too... But everything seems correct. Field is in T, converted from nano-ohms to Ohms, time in s, radius in m...

Answer 41

we have to divide by \(R\) not by \(\rho\)

Answer 42

please wait a I moment, I want to do a little research on the terrestrial magnetic field

Answer 43

Well, I suppose I should leave it in terms of S (cross sectional area of the collar) then. Normally our professor would specify if we need to leave it in terms of a variable but I don't see another way to do it.

Answer 44

In that case the formula for the requested current, is:
\[I = \frac{{rB}}{{2\rho \Delta t}}s\]

Answer 45

\[\Huge I = \frac{{rB}}{{2\rho \Delta t}}s\]

Answer 46

yes, some info seems to be missing here

a fairly simple approach is to take Faraday's law \(\large \mathcal{E} = -{{d\Phi} \over dt} \ \) noting that \(|\Phi| = B . A\) and \(A = \pi r^2 \cos \theta\) ...... where \(\theta\) is the angle between (A) the vector that is normal to the face of the ring and (B) the field ...... which angle varies from \( 0 \; \text{ to } \; \pi/2\) over the 0.2s.

so \(\large \mathcal{E} = -{{d\Phi} \over dt} \ = -B \pi R^2 {{d[ \cos \theta]} \over dt} = B \pi R^2\ \sin \theta \, \dot \theta \)

assuming \(\dot \theta = const. = \dfrac{\pi/2}{0.2} = \dfrac{5 \pi}{2}\) and inserting resistance as \(\dfrac{\rho 2 \pi R}{a}\) where a is the cross sectional area of the gold ring, seems that to convert this to a current using the known resistivity or conductivity of gold, you do need to know the radius of the ring.

\(\large I(\theta) = \dfrac{B \pi R^2\ \sin \theta \, \frac{5 \pi }{2} . a}{\rho 2 \pi R} = \dfrac{5\pi R B }{4 \rho}.a.\sin \theta\)

Answer 47

@IrishBoy123 I don't understand the significance and inclusion of \(\dot \theta = const. = \dfrac{\pi/2}{0.2} = \dfrac{5 \pi}{2}\) =/

Answer 48

it goes from perp to parallel in 0.2 sec.

so \(\theta\) goes from 0 to π/2 in that time

Answer 49

@Michele_Laino Ok so I get about 339.67*S amps. If we randomly choose that S has a radius of 1cm=0.01m, then we have an area of 3.14E-4.
(3.14E-4)(339.67)=0.107amps
Is this reasonable now? :P

Answer 50

Yes, I think so!