Question:medium

The magnetic field ((B)) inside a long solenoid carrying current ' (i) ' with an iron core is (( \mu_0 =) vacuum permeability, (\chi =) susceptibility)

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Ferromagnetic cores significantly increase the magnetic field of a solenoid because $\chi$ is very large.
Updated On: May 14, 2026
  • (\mu_0 ni(1 + \chi))
  • (\mu_0 ni^2(1 + \chi))
  • (\mu_0 ni\chi)
  • (\mu_0 ni(1 - \chi))
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
The magnetic field inside a solenoid changes when a material (like iron) is placed inside it. We need to express this new field in terms of the material's susceptibility \(\chi\).
Step 2: Key Formula or Approach:
1) Magnetic field in vacuum: \(B_0 = \mu_0 n i\).
2) Magnetic field in a medium: \(B = \mu n i = \mu_r \mu_0 n i\).
3) Relation between relative permeability and susceptibility: \(\mu_r = 1 + \chi\).
Step 3: Detailed Explanation:
The magnetic field strength \(H\) inside a long solenoid is \(H = n i\).
The total magnetic induction \(B\) inside the core material is given by:
\[ B = \mu H \]
where \(\mu\) is the absolute permeability of the core material.
We know that \(\mu = \mu_0 \mu_r\), where \(\mu_r\) is the relative permeability.
Substituting this into the expression for \(B\):
\[ B = \mu_0 \mu_r ni \]
From magnetic property relations, the relative permeability is related to magnetic susceptibility \(\chi\) by the equation:
\[ \mu_r = 1 + \chi \]
Therefore, the total magnetic field is:
\[ B = \mu_0 (1 + \chi) ni \]
Step 4: Final Answer:
The magnetic field is \(\mu_0 ni(1 + \chi)\).
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