MEC ChYU 1

Step 1: Determine the Axial Component ($B_x$)

The magnetic field on the axis of a solenoid extending from $x_1$ to $x_2$ is given by:

$$ B_{axis} = \frac{\mu_0 n I}{2} (\cos\theta_1 – \cos\theta_2) $$

For a semi-infinite solenoid positioned from $-\infty$ to $0$, the angles subtended by the ends at a point $x > 0$ on the axis are:

• $\theta_1 \to 0$ (for the end at $-\infty$), so $\cos\theta_1 = 1$.
• $\theta_2 = \theta$ (for the end at $0$), where $\cos\theta = \frac{x}{\sqrt{R^2 + x^2}}$.

Thus, the axial magnetic field $B_x$ at distance $x$ is:

$$ B_x(x) = \frac{\mu_0 n I}{2} \left[ 1 – \frac{x}{\sqrt{R^2 + x^2}} \right] $$

Step 2: Relate Axial and Radial Components using Gauss’s Law

Since magnetic monopoles do not exist, the net magnetic flux through any closed surface is zero ($\oint \vec{B} \cdot d\vec{A} = 0$).

Consider a small cylindrical “pillbox” of radius $r$ and length $dx$ coaxial with the solenoid, located at distance $x$ (as shown in the diagram above).

The total flux equation is:

$$ \Phi_{net} = \Phi_{out, axial} – \Phi_{in, axial} + \Phi_{radial} = 0 $$ $$ (B_x + dB_x)(\pi r^2) – B_x(\pi r^2) + B_r(2\pi r dx) = 0 $$

Simplifying this:

$$ \pi r^2 dB_x + 2\pi r dx B_r = 0 $$ $$ r dB_x + 2 dx B_r = 0 $$ $$ B_r = -\frac{r}{2} \frac{dB_x}{dx} $$

Step 3: Calculate the Derivative

Now we differentiate $B_x$ with respect to $x$ to find $B_r$.

Recall $B_x = \frac{\mu_0 n I}{2} \left[ 1 – x(R^2 + x^2)^{-1/2} \right]$.

$$ \frac{dB_x}{dx} = \frac{\mu_0 n I}{2} \left[ 0 – \frac{d}{dx}\left( \frac{x}{\sqrt{R^2+x^2}} \right) \right] $$

Using the quotient rule (or product rule):

$$ \frac{d}{dx}\left( \frac{x}{\sqrt{R^2+x^2}} \right) = \frac{1 \cdot \sqrt{R^2+x^2} – x \cdot \frac{1}{2}(R^2+x^2)^{-1/2}(2x)}{R^2+x^2} $$ $$ = \frac{(R^2+x^2) – x^2}{(R^2+x^2)^{3/2}} = \frac{R^2}{(R^2+x^2)^{3/2}} $$

So,

$$ \frac{dB_x}{dx} = -\frac{\mu_0 n I}{2} \frac{R^2}{(R^2+x^2)^{3/2}} $$

Step 4: Determine $B_r$ and Final Vector Expression

Substitute the derivative back into our expression for $B_r$:

$$ B_r = -\frac{r}{2} \left( -\frac{\mu_0 n I R^2}{2(R^2+x^2)^{3/2}} \right) $$ $$ B_r = \frac{\mu_0 n I}{2} \left[ \frac{r R^2}{2(R^2+x^2)^{3/2}} \right] $$

Combining the axial ($\hat{e}_x$) and radial ($\hat{e}_r$) components, the total magnetic field vector is: