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Physics Question 39 – JEE-MAIN 2025

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The percentage increase in magnetic field (B) when space within a current carrying solenoid is filled with magnesium (magnetic susceptibility χMg=1.2×10−5) is :

Understand how the magnetic field inside a solenoid changes when a material is introduced compared to when it's in a vacuum.

🥷
Ninja Strategy•Power of Ten Check

Recognize that percentage increase involves multiplying by 100 (or 102), so the power of 10 in the answer should be two greater than the power of 10 in the susceptibility. Only option 2 has the correct power of 10−3%.

Step 1: Relate Magnetic Field to Susceptibility✦ Active

The magnetic field inside a solenoid in vacuum is B0=μ0nI. When filled with a material of magnetic susceptibility χ, the magnetic field becomes B=μnI, where μ=μ0(1+χ) is the permeability of the material. Thus, B=μ0(1+χ)nI=B0(1+χ).

Step 2: Calculate Percentage Increase○ Expand

The percentage increase in the magnetic field is given by:

Percentage Increase=B−B0B0×100%

Substitute B=B0(1+χ) into the formula:

Percentage Increase=B0(1+χ)−B0B0×100%=B0χB0×100%=χ×100%
💡 Teacher's Secret Hint

Remember to multiply by 100 to convert the fractional increase into a percentage.

Step 3: Substitute Given Value and Final Calculation○ Expand

Given the magnetic susceptibility χMg=1.2×10−5. Substitute this value:

Percentage Increase=(1.2×10−5)×100%=1.2×10−3%

Convert 1.2 to a fraction 1210=65:

Percentage Increase=65×10−3%

This matches option 2.

💡 Teacher's Secret Hint

Pay attention to the powers of 10 and fractional conversions to match the options.

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