Magnetic Field Increase in Solenoid with Core

❓ Question

The percentage increase in magnetic field $B$ when the space within a current-carrying solenoid is filled with magnesium, given that the magnetic susceptibility of magnesium is

$${\chi }_{\text{Mg}}=1.2\times {10}^{-5},$$

is equal to ?

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🖼️ Question Image

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✍️ Short Solution

This is a direct concept-based JEE question.
No lengthy derivation — just one clean relation between magnetic field and susceptibility.

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🔹 Step 1 — Magnetic field inside a solenoid (key idea 🔥)**

For a long solenoid:

• In air (or vacuum):

$$B_0={\mu }_{0}nI$$

• When filled with a magnetic material:

$$B={\mu }_0(1+\chi )nI$$

Here:

• $\chi$ = magnetic susceptibility of the material

• $n$ = turns per unit length

• $I$ = current

📌 The geometry and current remain the same — only permeability changes.

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🔹 Step 2 — Change in magnetic field

Increase in magnetic field:

$$\mathrm{\Delta }B=B-B_0={\mu }_{0}nI\chi$$

So,

$$\frac{\mathrm{\Delta }B}{B_0}=\chi$$

👉 This is the golden result.

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🔹 Step 3 — Percentage increase formula

$$\mathrm{\%}\text{ increase in }B=\chi \times 100$$

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🔹 Step 4 — Substitute given value

Given:

$${\chi }_{\text{Mg}}=1.2\times {10}^{-5}$$

So,

$$\mathrm{\%}\text{ increase}=1.2\times {10}^{-5}\times 100=1.2\times {10}^{-3}\mathrm{\%}$$

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🔹 Step 5 — Physical interpretation (JEE insight 🧠)**

• Magnesium is a paramagnetic material

• Paramagnets have very small positive susceptibility

• Hence, the magnetic field increases very slightly

📌 This explains why the percentage increase is extremely small.

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✅ Final Answer

$$\boxed{{\displaystyle 1.2\times {10}^{-3}\mathrm{\%}}}$$