Calculating uncertainty in a experiment

[bobred]

Homework Statement

Finding uncertainty in magnetic field strength. $N=2$, $i=1.00\pm0.005$ A, $R=0.1$ m, $\tan \theta=0.49\pm0.008$

Homework Equations

The equation

$B_{coil}=\frac{\mu_0 N i}{2 R \tan \theta}$

The Attempt at a Solution

I am assuming that the radius R and number of turns N are exact.
I know how to find the error when the equation is X=a/b
Should I find this error and scale it by
$\frac{\mu_0 N}{2 R}$?

 

[mathfeel]

For multivariable formula $y = f(x_1, x_2, \dots, x_n)$: $\Delta y^2 = \sqrt{\sum_{j=1}^{n} \left( \frac{\partial f}{\partial x_j} \Delta x_j \right)^2}$

It looks tedious, but try to work out and simplify the algebra before you plug in number.

 

[PeterO]

Homework Statement

Finding uncertainty in magnetic field strength. $N=2$, $i=1.00\pm0.005$ A, $R=0.1$ m, $\tan \theta=0.49\pm0.008$

Homework Equations

The equation

$B_{coil}=\frac{\mu_0 N i}{2 R \tan \theta}$

The Attempt at a Solution

I am assuming that the radius R and number of turns N are exact.
I know how to find the error when the equation is X=a/b
Should I find this error and scale it by
$\frac{\mu_0 N}{2 R}$?

Your proposal sounds fine to me.

 

[bobred]

Hi

Sorted it out, my problem, I mis-read the formula

$\frac{\delta X}{X}=\sqrt{(\frac{\delta A}{A})^2+(\frac{\delta B}{B})^2}$

For $\delta X$ I needed to have

$\delta X=\sqrt{(\frac{\delta A}{A})^2+(\frac{\delta B}{B})^2} X$

Silly mistake
Thanks

 

[rwisz]

Yes, you are on the right track. In order to calculate the uncertainty in the magnetic field strength, you will need to use the error propagation formula:

\sigma_B = \sqrt{\left(\frac{\partial B}{\partial i}\sigma_i\right)^2 + \left(\frac{\partial B}{\partial \theta}\sigma_\theta\right)^2}

where \sigma_B is the uncertainty in B, \sigma_i is the uncertainty in i, and \sigma_\theta is the uncertainty in \theta.

In this case, the partial derivatives are:

\frac{\partial B}{\partial i} = \frac{\mu_0 N}{2 R \tan \theta}

\frac{\partial B}{\partial \theta} = -\frac{\mu_0 N i}{2 R \tan^2 \theta}

Plugging these values into the error propagation formula, we get:

\sigma_B = \sqrt{\left(\frac{\mu_0 N}{2 R \tan \theta}\sigma_i\right)^2 + \left(-\frac{\mu_0 N i}{2 R \tan^2 \theta}\sigma_\theta\right)^2}

= \sqrt{\left(\frac{\mu_0 N}{2 R \tan \theta}\right)^2\sigma_i^2 + \left(\frac{\mu_0 N i}{2 R \tan^2 \theta}\right)^2\sigma_\theta^2}

= \frac{\mu_0 N}{2 R \tan \theta}\sqrt{\sigma_i^2 + \left(\frac{i}{\tan \theta}\sigma_\theta\right)^2}

= \frac{\mu_0 N}{2 R \tan \theta}\sqrt{\sigma_i^2 + \left(\frac{i\sigma_\theta}{\cos \theta}\right)^2}

= \frac{\mu_0 N}{2 R \tan \theta}\sqrt{\sigma_i^2 + \left(\frac{i\sigma_\theta}{\sqrt{1-\tan^2 \theta}}\right)^2}

= \frac{\mu_0 N}{2 R \tan \theta}\sqrt{\sigma_i^2 + \left(\frac{i\sigma_\theta}{\sqrt{1-\tan^2 \theta}}\right)^2}

= \frac{\mu_0 N}{2 R \