Question

The magnetic field intensity 'H' at the centre of a long solenoid having 'n' turns per unit length and carrying a current 'I', when no material is kept in it, is

• A. $\frac{1}{nI}$
• B. $\frac{n}{I}$
• C. $nI$
• D. $n^2 I$

Hint:

Always keep your unit relations straight: Magnetic induction $B$ depends on the medium properties ($\mu_0$), whereas magnetic intensity $H$ is independent of the medium and depends strictly on the current configuration geometry ($nI$)!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the magnetic field intensity ($H$) at the core center of a long vacuum-core solenoid with $n$ turns per unit length carrying a steady current $I$.

Step 2: Detailed Explanation:
Let's distinguish between magnetic induction ($B$) and magnetic field intensity ($H$):

• According to Ampere's Circuital Law, the total magnetic induction field ($B$) inside an ideal long air-core solenoid is: $$ B = \mu_0 nI $$ where $\mu_0$ is the permeability of free space.

• The magnetic field intensity vector $\vec{H}$ represents the external magnetizing field strength and is fundamentally related to $\vec{B}$ in a vacuum medium by the formula: $$ B = \mu_0 H \implies H = \frac{B}{\mu_0} $$
Substituting the expression for $B$ into our intensity relation: $$ H = \frac{\mu_0 nI}{\mu_0} = nI $$

Step 3: Final Answer:
The magnetic field intensity $H$ at the center of the solenoid is $nI$, which corresponds to option (C).