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JEE Main 2025 (29 Jan, Shift 2) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2025 (29 Jan, Shift 2)

Q14JEE Main 2025 · 29 JAN · SHIFT 2Easy

Ampere's Law, Cylinders, Solenoid & Toroid

The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9×10−4 Tesla2.9 \times 10^{-4}\text{ Tesla}. If the solenoid carries a current of 0.29 A, then the length of the solenoid is _______ πcm\pi\text{cm}.

Numerical answer type. Work it out, then open the answer.

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Answer: 8

  1. B=μ0NIL⇒L=μ0NIBB = \dfrac{\mu_0 N I}{L} \Rightarrow L = \dfrac{\mu_0 N I}{B} (radius not needed).
  2. L=4π×10−7×200×0.292.9×10−4=4π×10−7×2×105L = \dfrac{4\pi \times 10^{-7} \times 200 \times 0.29}{2.9 \times 10^{-4}} = 4\pi \times 10^{-7} \times 2 \times 10^{5}.
  3. L=0.08πL = 0.08\pi m =8π= 8\pi cm.
  4. Answer: 8
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