Use superposition for the cavity

A current-carrying cylinder with an off-center cavity can be treated as:

• a full solid cylinder of radius $a$ carrying uniform current density $J$, and
• a smaller cylinder (the cavity) of radius $a/2$ carrying current density $-J$ at the cavity location.

This cancels current in the cavity region.

Magnetic field inside a uniformly current-filled infinite cylinder

For a point at distance $r$ from the axis of a cylinder carrying uniform current density $J$, $B=\frac{\mu_0 J r}{2}$ and the direction is tangential by right-hand rule.

Geometry from the figure

The main cylinder has diameter $2a$, so its radius is $R=a.$ The cavity has diameter $a$, so its radius is $r_c=\frac a2.$

Since the cavity is tangent internally to the outer cylinder, the distance between the two axes is $OO'=a-\frac a2=\frac a2.$

From the usual figure for this standard problem, point $P$ is the point on the outer surface diametrically opposite the cavity side. Hence:

• distance of $P$ from main axis $O$: $OP=a$
• distance of $P$ from cavity axis $O'$: $O'P=a+\frac a2=\frac{3a}{2}$

Field due to the full cylinder at $P$

Since $P$ lies on the surface of the full cylinder, $B_1=\frac{\mu_0 J a}{2}.$

Field due to the negative-current cavity cylinder at $P$

Here $P$ is outside the small cylinder because $O'P=3a/2>a/2$.

Total current of the small cylinder is $I_c=J\cdot \pi\left(\frac a2\right)^2=\frac{J\pi a^2}{4}.$

So field at distance $3a/2$ is

Direction of the two fields

At point $P$, both fields are along the same tangential direction in magnitude construction, but because the cavity cylinder carries current density $-J$, its field is opposite to what a positive current would produce at that point. Hence the net field is

Compare with the given form

Given $B=\frac{N}{12}\mu_0 aJ,$ so $N=5.$

Comparison with stored answer

Derived answer: $5$

Stored correct answer: $5$

They agree.