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JEE Main 2020 (5 Sep, Shift 2) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2020 (5 Sep, Shift 2)

Q49JEE Main 2020 · 5 SEP · SHIFT 2Medium

Ampere's Law, Cylinders, Solenoid & Toroid

An iron rod of volume 10–3 m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be:

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Answer: (a)

  1. H=nI=1000×0.5=500H = nI = 1000 \times 0.5 = 500 A/m (10 turns/cm = 1000 turns/m).
  2. Magnetisation M=(μr−1)H=999×500≈5×105M = (\mu_r - 1) H = 999 \times 500 \approx 5 \times 10^{5} A/m.
  3. Magnetic moment m=MV=5×105×10−3=5×102m = MV = 5 \times 10^{5} \times 10^{-3} = 5 \times 10^{2} A m2^2.
  4. So the answer is (a) 5×1025 \times 10^2 A m2^2.
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