Magnetic Field at Point near Cylindrical Cavity in Current Conductor

To find the magnitude of the magnetic field at point $P$, we can use the principle of superposition.

1. Geometry and Coordinate System

Let the center of the main cylinder $O$ be the origin $(0, 0)$.

• The main cylinder has a diameter of $2a$, which gives it a radius of $R = a$.
• Point $P$ is located on the outer surface of the main cylinder on the opposite side of the cavity, at coordinates $(-a, 0)$.
• The cylindrical cavity has a diameter of $a$, meaning its radius is $r = \frac{a}{2}$. Its center $O_c$ lies on the line connecting $P$ and $O$, at coordinates $\left(\frac{a}{2}, 0\right)$.

Using superposition, the configuration can be treated as the combination of:

1. A solid cylinder of radius $R = a$ carrying a uniform current density $+J \hat{k}$.
2. A smaller cylinder corresponding to the cavity, of radius $r = \frac{a}{2}$, centered at $O_c$, carrying a uniform current density $-J \hat{k}$.

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2. Magnetic Field due to the Solid Cylinder ($\mathbf{B}_1$)

Point $P(-a, 0)$ lies on the boundary of the solid cylinder ($r_1 = a$). By Ampere's Circuital Law:

$\oint \mathbf{B}_1 \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}}$

$B_1 (2\pi a) = \mu_0 \left( J \cdot \pi a^2 \right)$

$B_1 = \frac{\mu_0 J a}{2}$

Assuming the current density $+J$ flows out of the page ($+\hat{k}$), by the right-hand rule, the magnetic field at $P(-a, 0)$ points in the $-\hat{j}$ direction:

$\mathbf{B}_1 = -\frac{\mu_0 J a}{2} \hat{j}$

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3. Magnetic Field due to the Cavity Cylinder ($\mathbf{B}_2$)

The distance $d$ from the center of the cavity $O_c\left(\frac{a}{2}, 0\right)$ to point $P(-a, 0)$ is:

$d = a + \frac{a}{2} = \frac{3a}{2}$

Since $d = \frac{3a}{2} > \frac{a}{2}$, point $P$ lies outside the cavity region. The magnetic field at $P$ due to this smaller cylinder carrying total current $I_{\text{cav}} = -J \left[\pi \left(\frac{a}{2}\right)^2\right]$ is:

$B_2 = \frac{\mu_0 |I_{\text{cav}}|}{2\pi d} = \frac{\mu_0 J \pi \left(\frac{a}{2}\right)^2}{2\pi \left(\frac{3a}{2}\right)} = \frac{\mu_0 J \frac{a^2}{4}}{3a} = \frac{\mu_0 J a}{12}$

Since this region carries a current in the $-\hat{k}$ direction, the right-hand rule gives the direction of $\mathbf{B}_2$ at $P$ along the $+\hat{j}$ direction:

$\mathbf{B}_2 = +\frac{\mu_0 J a}{12} \hat{j}$

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4. Net Magnetic Field at Point $P$

By superposition, the net magnetic field $\mathbf{B}_{\text{net}}$ is:

$\mathbf{B}_{\text{net}} = \mathbf{B}_1 + \mathbf{B}_2 = \left( -\frac{\mu_0 J a}{2} + \frac{\mu_0 J a}{12} \right) \hat{j} = -\frac{5}{12} \mu_0 J a \hat{j}$

The magnitude of the magnetic field is:

$|\mathbf{B}_{\text{net}}| = \frac{5}{12} \mu_0 J a$

Comparing this with the given expression $\frac{N}{12} \mu_0 a J$, we get:

$N = 5$