A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. B

Q: Why do we use $-J$ for the cavity?

A cavity means there is no material and no current in that region. Mathematically, adding a current density of $+J$ and $-J$ in the same region results in zero net current, which perfectly models the empty cavity while allowing us to use standard formulas for complete cylinders.

Q: How did we get the distance $3a/2$ for the cavity?

Point $P$ is on the left edge of the large cylinder, at a distance $a$ from the main center $O$. The cavity's center $O'$ is at a distance $a/2$ to the right of $O$. Therefore, the total distance from $O'$ to $P$ is $a + a/2 = 3a/2$.

Q: Why do we subtract the magnetic fields?

The magnetic field is a vector. The solid cylinder's current ($+J$) produces a field pointing downwards at $P$, while the cavity's reverse current ($-J$) produces a field pointing upwards at $P$. Since they are in opposite directions, their magnitudes subtract.

Q: Can we use Ampere's law directly on the system with the cavity?

No, because the system with the cavity lacks the necessary cylindrical symmetry. Ampere's law is only easily solvable when the magnetic field is constant along the chosen Amperian loop, which requires high symmetry.

Q: What is the direction of the net magnetic field?

Since the magnetic field from the solid cylinder ($B_T$) is larger than the field from the cavity ($B_C$), the net magnetic field will point in the direction of $B_T$, which is downwards (assuming the current $J$ flows into the page).

Question

A cylindrical cavity of diameter $a$ exists inside a cylinder of diameter $2a$ as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density $J$ flows along the length. If the magnitude of the magnetic field at the point $P$ is given by $\frac{N}{12}\mu_0 aJ$, then the value of $N$ is

Sigma Percentile · Accepted Answer

The concept of a "cavity" in physics often feels like a trick question. How do you calculate the magnetic field of something that *isn't* there? The secret lies in a beautiful mathematical trick: the **Principle of Superposition**.

Instead of dealing with a complex shape with a hole in it, we can imagine the system as two complete, simple shapes overlapping each other.

### Analyzing the Geometry

Before diving into the physics, let's map out the geometry of our system. We have a large cylinder with a diameter of $2a$, which means its radius is $R_1 = a$. Let's place its center at the origin, $O$.

Inside it, there is a cylindrical cavity of diameter $a$, meaning its radius is $R_2 = a/2$. Looking at the figure, the cavity touches the right edge of the large cylinder and its center $O'$ is located at a distance of $a/2$ to the right of the main center $O$.

We need to find the magnetic field at point $P$, which is on the far left edge of the large cylinder.
- The distance from the main center $O$ to point $P$ is $r_1 = a$.
- The distance from the cavity's center $O'$ to point $P$ is $r_2 = a + a/2 = \frac{3a}{2}$.

### The Superposition Principle

To find the net magnetic field at $P$, we treat the system as a combination of two parts:
1. A complete, solid cylinder of radius $a$ carrying a uniform current density $+J$.
2. A smaller solid cylinder (representing the cavity) of radius $a/2$ carrying a reverse current density $-J$.

When these two overlap, the $+J$ and $-J$ in the cavity region cancel out perfectly to zero, leaving us with the exact physical situation described in the problem!

### Magnetic Field of the Solid Cylinder

Let's calculate the magnetic field $\mathbf{B}_T$ produced by the complete solid cylinder at point $P$. First, we need the total current $I_T$ flowing through it.
$$I_T = J \cdot 	ext{Area} = J(\pi a^2)$$

Using Ampere's Circuital Law, the magnetic field at a distance $r$ from the center of a long straight wire is $B = \frac{\mu_0 I}{2\pi r}$. For point $P$, the distance is $r_1 = a$.
$$B_T = \frac{\mu_0 (J \pi a^2)}{2\pi a} = \frac{\mu_0 J a}{2}$$

By the right-hand grip rule, if the current is flowing into the page, this magnetic field at point $P$ (which is to the left of the center) will point straight down.

### Magnetic Field of the Cavity

Next, we calculate the magnetic field $\mathbf{B}_C$ produced by the "negative" current in the cavity region. The current $I_C$ here is:
$$I_C = J \cdot \pi \left(\frac{a}{2}ight)^2 = \frac{J \pi a^2}{4}$$

The distance from the cavity's center $O'$ to point $P$ is $r_2 = \frac{3a}{2}$. Applying Ampere's law again:
$$B_C = \frac{\mu_0 I_C}{2\pi r_2} = \frac{\mu_0 \left(\frac{J \pi a^2}{4}ight)}{2\pi \left(\frac{3a}{2}ight)}$$

Simplifying this expression:
$$B_C = \frac{\mu_0 J a}{12}$$

Since this current is flowing in the reverse direction (out of the page), the magnetic field it produces at point $P$ will point straight up.

### The Final Calculation

We now have two magnetic fields at point $P$: $\mathbf{B}_T$ pointing down and $\mathbf{B}_C$ pointing up. The net magnetic field $B_{net}$ is their difference.
$$B_{net} = B_T - B_C$$
$$B_{net} = \frac{\mu_0 J a}{2} - \frac{\mu_0 J a}{12}$$

To subtract these, we find a common denominator of 12:
$$B_{net} = \frac{6\mu_0 J a}{12} - \frac{\mu_0 J a}{12} = \frac{5\mu_0 J a}{12}$$

The problem states that the magnitude of the magnetic field at $P$ is given by $\frac{N}{12}\mu_0 aJ$. By comparing our result with this expression, we can clearly see that:
**$N = 5$**

Enter Numerical Value:

Visualized Solution

$G eo m e t r y o f t h e S y s t e m$

$P r in c i pl e o f S u p er p os i t i o n$

$M a g n e t i c F i e l d d u e t o S o l i d C y l in d er$

$M a g n e t i c F i e l d d u e t o C a v i t y$

$N e t M a g n e t i c F i e l d$

$F in d in g t h e V a l u e o f N$

The Sigma Insight: Ampere's Circuital Law

Analyzing the Geometry

The Superposition Principle

Magnetic Field of the Solid Cylinder

Magnetic Field of the Cavity

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Geometry of the System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Principle of Superposition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field due to Solid Cylinder

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field due to Cavity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Magnetic Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Value of N