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Magnetic Properties of Matter question

2025 · 7 Apr · Shift 1 · Q52
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  5. /2025 · 7 Apr · Shift 1 · Q52

Magnetic Properties of Matter question

2025 · 7 Apr · Shift 1 · Q52

JEE MainPhysicsMagnetic Properties of MatterMCQ+4 / −1
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\chi_{\mathrm{Mg}}=1.2 \times 10^{-5}χMg​=1.2×10−5 ) is :
  1. A
    53×10−5%\frac{5}{3} \times 10^{-5} \%35​×10−5%
  2. B
    56×10−4%\frac{5}{6} \times 10^{-4} \%65​×10−4%
  3. C
    56×10−5%\frac{5}{6} \times 10^{-5} \%65​×10−5%
  4. D
    65×10−3%\frac{6}{5} \times 10^{-3} \%56​×10−3%
View written solutionFree

Correct answer: D

  1. Magnetic field inside a solenoid without material

For an empty solenoid, B0=μ0nIB_0=\mu_0 n IB0​=μ0​nI where nnn is number of turns per unit length.

  1. When a magnetic material is inserted

If the solenoid is filled with a material of magnetic susceptibility χ\chiχ, then μ=μ0(1+χ)\mu = \mu_0(1+\chi)μ=μ0​(1+χ) So the magnetic field becomes B=μnI=μ0(1+χ)nI=B0(1+χ)B = \mu n I = \mu_0(1+\chi)nI = B_0(1+\chi)B=μnI=μ0​(1+χ)nI=B0​(1+χ)

  1. Increase in magnetic field

Hence, ΔB=B−B0=B0χ\Delta B = B-B_0 = B_0\chiΔB=B−B0​=B0​χ Therefore percentage increase is ΔBB0×100=χ×100\frac{\Delta B}{B_0}\times 100 = \chi \times 100B0​ΔB​×100=χ×100

  1. Substitute for magnesium

Given, χMg=1.2×10−5\chi_{\mathrm{Mg}} = 1.2\times 10^{-5}χMg​=1.2×10−5 So, % increase=1.2×10−5×100\%\text{ increase} = 1.2\times 10^{-5}\times 100% increase=1.2×10−5×100 =1.2×10−3%=1.2\times 10^{-3}\%=1.2×10−3%

Now, 1.2×10−3=65×10−31.2\times 10^{-3} = \frac{6}{5}\times 10^{-3}1.2×10−3=56​×10−3 Thus, % increase=65×10−3%\%\text{ increase} = \frac{6}{5}\times 10^{-3}\%% increase=56​×10−3%

  1. Match with options

This corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer is D, which matches our result.

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