MEC CYU 1

Problem Statement Analysis:

We need to find the magnetic induction vector $\vec{B}$ at a point $P$ outside a semi-infinite solenoid. The point is at an axial distance $x$ from the end face and a small radial distance $r$ from the axis ($r \ll R$).

We will determine the axial component $B_x$ using the standard formula for a point on the axis and then derive the radial component $B_r$ using Gauss’s Law for magnetism ($\nabla \cdot \vec{B} = 0$).

Step 1: Axial Component ($B_x$)

For a semi-infinite solenoid (extending from $-\infty$ to $0$), the magnetic field at a point on the axis at a distance $x$ from the end face is given by:

$$ B_{axis} = \frac{\mu_0 n I}{2} (1 – \cos\alpha) $$

Where $\alpha$ is the semi-vertical angle subtended by the rim of the solenoid end at the point $x$. From the geometry:

$$ \cos\alpha = \frac{x}{\sqrt{R^2 + x^2}} $$

Thus, the axial component is:

$$ B_x(x, 0) = \frac{\mu_0 n I}{2} \left( 1 – \frac{x}{\sqrt{R^2 + x^2}} \right) $$

Since we are considering a point very close to the axis ($r \ll R$), the variation of $B_x$ with $r$ is negligible to the first order. We assume $B_x(x, r) \approx B_x(x, 0)$.

Step 2: Radial Component ($B_r$)

We use Gauss’s Law for magnetism, $\nabla \cdot \vec{B} = 0$. In cylindrical coordinates with azimuthal symmetry ($\frac{\partial B_\phi}{\partial \phi} = 0$), this is:

$$ \frac{\partial B_x}{\partial x} + \frac{1}{r} \frac{\partial}{\partial r} (r B_r) = 0 $$ $$ \frac{1}{r} \frac{\partial}{\partial r} (r B_r) = – \frac{\partial B_x}{\partial x} $$

We can also get it by taking a pillbox gaussian surface (instead of partial differentiation)

First, calculate the derivative of $B_x$ with respect to $x$:

$$ \frac{\partial B_x}{\partial x} = \frac{\mu_0 n I}{2} \frac{d}{dx} \left( 1 – x(R^2 + x^2)^{-1/2} \right) $$ $$ \frac{\partial B_x}{\partial x} = \frac{\mu_0 n I}{2} \left[ – \left( (R^2+x^2)^{-1/2} + x \cdot \left(-\frac{1}{2}\right)(R^2+x^2)^{-3/2} \cdot 2x \right) \right] $$ $$ \frac{\partial B_x}{\partial x} = -\frac{\mu_0 n I}{2} \left[ \frac{1}{\sqrt{R^2+x^2}} – \frac{x^2}{(R^2+x^2)^{3/2}} \right] $$ $$ \frac{\partial B_x}{\partial x} = -\frac{\mu_0 n I}{2} \left[ \frac{R^2+x^2 – x^2}{(R^2+x^2)^{3/2}} \right] = -\frac{\mu_0 n I R^2}{2(R^2+x^2)^{3/2}} $$

Substitute this back into the divergence equation:

$$ \frac{1}{r} \frac{\partial}{\partial r} (r B_r) = – \left( -\frac{\mu_0 n I R^2}{2(R^2+x^2)^{3/2}} \right) = \frac{\mu_0 n I R^2}{2(R^2+x^2)^{3/2}} $$

Now, integrate with respect to $r$:

$$ r B_r = \int \frac{\mu_0 n I R^2}{2(R^2+x^2)^{3/2}} r dr $$ $$ r B_r = \frac{\mu_0 n I R^2}{2(R^2+x^2)^{3/2}} \cdot \frac{r^2}{2} $$ $$ B_r = \frac{\mu_0 n I r R^2}{4(R^2+x^2)^{3/2}} $$

Final Answer

Combining the components, the magnetic induction vector is:

Note: The sign in the axial term corresponds to a point outside the solenoid where the field strength decreases as $x$ increases.