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JEE Main 2006 (AIEEE) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2006 (AIEEE)

Q58JEE Main 2006 · AIEEEEasy

Ampere's Law, Cylinders, Solenoid & Toroid

A long solenoid has 200200 turns per cmcm and carries a current i.i. The magnetic field at its center is 6.28×10−2 Weber/m2.6.28 \times 10^{-2}\ Weber/m^2. Another long solenoid has 100100 turns per cmcm and it carries a current i3\frac{i}{3}. The value of the magnetic field at its center is

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Answer: (a)

  1. B=μ0nI∝nIB = \mu_0 n I \propto nI.
  2. B2B1=100200×13=16\dfrac{B_2}{B_1} = \dfrac{100}{200} \times \dfrac{1}{3} = \dfrac{1}{6}.
  3. B2=6.28×10−26=1.05×10−2B_2 = \dfrac{6.28 \times 10^{-2}}{6} = 1.05 \times 10^{-2} Wb/m2^2.
  4. So the answer is (a).
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