Question

Figure shows the cross-sectional view of the hollow cylindrical conductor with inner radius R and outer radius '2R, cylinder carrying uniformly distributed current along it's length in metal. The magnetic induction at point 'P' at a distance $\frac{3R}{2}$ from the axis of the cylinder will be

$i \longrightarrow \pi(2R)^2$

$\frac{i}{\pi(2R)^2}$

• A. Zero
• B. $\frac{5\mu_0i}{72\pi R}$
• C. $\frac{7\mu_0i}{18\pi R}$
• D. $\frac{5\mu_0i}{36\pi R}$

Answer 1

Answer: (D)

(4) $\frac{5\mu_0i}{36\pi R}$

Answer 2

To determine the magnetic induction at point 'P' at a distance $\frac{3R}{2}$ from the axis of the hollow cylindrical conductor, we use Ampere's Law.

1. Identify the region:
   The conductor has an inner radius $R$ and an outer radius $2R$. The point 'P' is at a distance $r = \frac{3R}{2}$ from the axis. Since $R < \frac{3R}{2} < 2R$, point 'P' lies within the material of the conductor.

2. Apply Ampere's Law:
   For a symmetrical current distribution, the magnetic field $\vec{B}$ is tangential to a circular Amperian loop concentric with the cylinder. Ampere's Law states:

   $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$

   For a circular Amperian loop of radius $r$:

   $$B (2\pi r) = \mu_0 I_{enc}$$

3. Calculate the enclosed current ($I_{enc}$):
   The total current $i$ is uniformly distributed over the cross-sectional area of the conductor.
   The total area of the conductor's cross-section is $A_{total} = \pi (2R)^2 - \pi R^2 = \pi (4R^2 - R^2) = 3\pi R^2$.
   The current density $J$ is:

   $$J = \frac{i}{A_{total}} = \frac{i}{3\pi R^2}$$

   The Amperian loop at radius $r = \frac{3R}{2}$ encloses current flowing through the annular region from $R$ to $r$. The area of this enclosed region is:

   $$A_{enc} = \pi r^2 - \pi R^2 = \pi \left(\frac{3R}{2}\right)^2 - \pi R^2$$ $$A_{enc} = \pi \left(\frac{9R^2}{4}\right) - \pi R^2 = \pi R^2 \left(\frac{9}{4} - 1\right) = \pi R^2 \left(\frac{5}{4}\right)$$

   The enclosed current $I_{enc}$ is the current density multiplied by the enclosed area:

   $$I_{enc} = J \times A_{enc} = \left(\frac{i}{3\pi R^2}\right) \times \left(\pi R^2 \frac{5}{4}\right)$$ $$I_{enc} = \frac{5i}{12}$$

4. Substitute into Ampere's Law and solve for B:

   $$B (2\pi r) = \mu_0 I_{enc}$$

   Substitute $r = \frac{3R}{2}$ and $I_{enc} = \frac{5i}{12}$:

   $$B \left(2\pi \frac{3R}{2}\right) = \mu_0 \left(\frac{5i}{12}\right)$$ $$B (3\pi R) = \frac{5\mu_0 i}{12}$$ $$B = \frac{5\mu_0 i}{12 \times 3\pi R}$$ $$B = \frac{5\mu_0 i}{36\pi R}$$

This result matches option (4).

The final answer is $\boxed{\text{(4)}}$.

Explanation of the solution:
The magnetic induction at a point inside a hollow cylindrical conductor is found using Ampere's Law. For a point at radius $r$ ($R_{inner} < r < R_{outer}$), the magnetic field is $B(2\pi r) = \mu_0 I_{enc}$. The enclosed current $I_{enc}$ is calculated by multiplying the current density $J = \frac{I_{total}}{A_{total}}$ by the area of the conductor enclosed by the Amperian loop, $A_{enc} = \pi r^2 - \pi R_{inner}^2$. Substituting $R_{inner}=R$, $R_{outer}=2R$, $r=\frac{3R}{2}$, and $I_{total}=i$, we get $A_{total} = 3\pi R^2$, $J = \frac{i}{3\pi R^2}$, and $A_{enc} = \pi (\frac{3R}{2})^2 - \pi R^2 = \frac{5\pi R^2}{4}$. Thus, $I_{enc} = \frac{i}{3\pi R^2} \times \frac{5\pi R^2}{4} = \frac{5i}{12}$. Plugging this into Ampere's Law: $B(2\pi \frac{3R}{2}) = \mu_0 \frac{5i}{12}$, which simplifies to $B(3\pi R) = \frac{5\mu_0 i}{12}$. Solving for $B$ yields $B = \frac{5\mu_0 i}{36\pi R}$.