Solenoid Exercise

As sir @Karan Chatrath has provided a very nice solution. I will just elaborate the solution. This problem can be considered as a biot-savart law problem because we have the exact information of solenoid structure.

Consider a point on the solenoid as such:

$\vec{r}_p = \cos{\theta} \hat{i} + \sin{\theta} \hat{j} + \left(-10+\frac{\theta}{100 \pi}\right)\hat{k}$

An arc length element tangential to the point is:

$d\vec{r}_p = \left(-\sin{\theta} \hat{i} + \cos{\theta} \hat{j} + \left(\frac{1}{100 \pi}\right)\hat{k}\right) \ d\theta$

Consider the point of interest at which field is to be computed to be:

$\vec{r}_{c1} = 0 \hat{i} + 0 \hat{j} +0\hat{k}$ $\vec{r}_{c2} = 2 \hat{i} + 0 \hat{j} +0\hat{k}$

In general, the point of interest will be denoted by $\vec{r}_{c}$ . The vector joining a point on the solenoid and the point of interest, and directed towards the point of interest, is:

$\vec{r} = \vec{r}_c - \vec{r}_p$

Now, Biot-Savart Law is applied:

$d\vec{B} = \frac{\mu_o I}{4 \pi}\left(\frac{d\vec{r}_p \times \vec{r}}{\lvert \vec{r} \rvert^3}\right)= f_x(\theta) \hat{i} + f_y(\theta) \hat{j} +f_z(\theta) \hat{k}$

$d\vec{B}$ has three components each of which is a function of $\theta$ only. This gives three integrals:

$B_x = \int_{0}^{2000 \pi} f_x(\theta) \ d\theta$
$B_y = \int_{0}^{2000 \pi} f_y(\theta) \ d\theta$ $B_z = \int_{0}^{2000 \pi} f_z(\theta) \ d\theta$

I have written a script of code to compute the answer for each case which turns out to be:

$\boxed{\frac{B_1}{B_2} \approx 201.1528}$

As for the bonus question, In the first case, the field magnitude turns out to be:

$B_1 \approx 49.75$

Which is almost close to the result produced by the standard formula which turns out to be $50$ . This is unlike the case where

$B_2 \approx 0.2473$

This result highly deviates from the standard formula prediction. It is surprising that just an offset of two units in the point of interest can produce such a different result. In my opinion, the standard formula holds true provided the test point is sufficiently and equally far from either end of the solenoid. I am yet to test this speculation, however.