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Numerical · Q26

Q.A solenoid electromagnet has 500500 turns wound uniformly over a length of 0.5 m0.5\ \text{m} and carries a current of 2 A2\ \text{A}. Calculate the magnetic field produced

(a) with an air core, and
(b) with a soft-iron core of relative permeability 10001000 filling the solenoid.
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Given N=500N=500, l=0.5 ml=0.5\ \text{m}, I=2 AI=2\ \text{A}, so turns per unit length n=N/l=500/0.5=1000 turns/mn=N/l=500/0.5=1000\ \text{turns/m}.

  1. Air core (μr≈1\mu_r\approx1):

    B0=μ0nI=(4π×10−7)×1000×2=4π×10−7×2000B_0 = \mu_0nI = (4\pi\times10^{-7})\times1000\times2 = 4\pi\times10^{-7}\times2000

    B0=8000π×10−7≈2.51×10−3 TB_0 = 8000\pi\times10^{-7} \approx 2.51\times10^{-3}\ \text{T}

  2. Soft-iron core (μr=1000\mu_r=1000): B=μrμ0nI=1000×B0=1000×2.51×10−3≈2.51 TB = \mu_r\mu_0nI = 1000\times B_0 = 1000\times2.51\times10^{-3} \approx 2.51\ \text{T} …
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