Magnetic flux density in the centre of a current carrying loop

[stobbz]

Homework Statement

A current carrying loop of wire is being monitored using a Hall-effect magnetic field probe placed at its centre. The loop is in air and has a radius of 0.5 m. It is required to monitor the magnetic flux in order to trip the circuit breaker if the current exceeds 100 A.

What level of B measured by the probe will correspond to this trip threshold?

Homework Equations

The Attempt at a Solution

Would I be right in saying the magnetic field strength is: H = I/2*PI*r = 31.83 A/m ?
I tried to determine B using: B = uH, this gave an incorrect answer, 40 micro Tesla.

The given answer is 1.26*10^-4 T.

Can anybody point me in the right direction?
Thanks.

 

[Mindscrape]

Hmm

$$B= I \int_0^{2 \pi} \frac{\mu_0}{4 \pi} \frac{ r d \theta}{r^2}$$
$$B = \frac{\mu_0 I}{2 R}$$

You've got the wrong formula. Were you supposed to derive it? I can't remember working with Biot Savart in calc physics 2, so hopefully it's okay I gave it do you.

Also, why are you using H? I don't see anything about the magnetic field being in a medium with a new permeability. Furthermore, this isn't magnetic flux density :p, sorry for being rude but terminology can be important.

 

[stobbz]

Ahh ok, thanks for the reply. I'd much rather you corrected me when I'm wrong :-)

I think I was supposed to derive it, but I have no idea where to start. This is causing me untold misery!

Thanks again.

 

[Mindscrape]

Well, here's how you derive it:

Biot Savart law is as follows

$$dB = \frac{I}{4\pi}\frac{dl \times \hat{r}}{r^2}$$

You know that the magnetic field at the center of the loop will be up, by symmetry the horizontal components cancel out, which means the cross product of the differential length with the direction of the field is 1 (they are perpendicular/orthogonal/however you want to say it). We want to integrate over an angle, and similar principle as how a speed along a circle is w*r, the angular differential for a length is $r d\theta$. We know what r is because we are in the very center it's simply the radius of the loop R (if we were further up we would of course have to apply pythagorean theorem).

$$dB = \frac{I}{4\pi}\frac{R d\theta}{R^2}$$

From there it's what I posted.

 

[rwisz]

I would first clarify the units being used in the given answer. It appears to be in Tesla (T), which is the SI unit for magnetic flux density. Therefore, the first step would be to convert the given answer of 1.26*10^-4 T to the corresponding units of A/m.

Next, I would use the equation B = uH to calculate the magnetic flux density at the centre of the current carrying loop. This equation relates the magnetic flux density (B) to the magnetic field strength (H) and the permeability of the medium (u). In this case, u can be assumed to be the permeability of air, which is approximately equal to 1.

Using the given value of H = 31.83 A/m, and assuming u = 1, the calculated value of B would be equal to 31.83*1 = 31.83 T. This is significantly different from the given answer, indicating that there may be a mistake in the given answer or in the calculations.

To further investigate, I would also consider the direction of the magnetic field at the centre of the loop. The magnetic field lines would be perpendicular to the plane of the loop, and their direction would depend on the direction of the current flow. If the current is flowing clockwise, the magnetic field at the centre of the loop would point outwards, and if the current is flowing counterclockwise, the magnetic field would point inwards.

In conclusion, to accurately determine the magnetic flux density at the centre of the current carrying loop, it is important to consider the correct units, the direction of the magnetic field, and the permeability of the medium. Further clarification of the given answer or the problem statement may be necessary to arrive at a correct solution.