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MHT CET 2026 (18 Apr, Shift 2) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from MHT CET 2026 (18 Apr, Shift 2)

Q2MHT CET 2026 · 18 APR · SHIFT 2Easy

Ampere's Law, Cylinders, Solenoid & Toroid

Two wires with current 3A and 1.5A are enclosed in a circular loop P. Third parallel wire with current 1A is situated outside the loop as shown. All the wires are perpendicular to the plane of the circular loop. The value of ∮B⃗⋅dl⃗\oint \vec{B} \cdot \vec{dl} around the loop is (μ0=\mu_0 = permeability of free space)

Figure

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Answer: (c)

  1. Ampere's law: ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}; the outside wire does not count.
  2. The 3 A and 1.5 A currents are opposite: Ienc=3−1.5=1.5I_{enc} = 3 - 1.5 = 1.5 A.
  3. ∮B⃗⋅dl⃗=1.5 μ0\oint \vec{B}\cdot d\vec{l} = 1.5\,\mu_0
  4. So the answer is (c).
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