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MHT CET 2026 (20 Apr, Shift 2) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from MHT CET 2026 (20 Apr, Shift 2)

Q1MHT CET 2026 · 20 APR · SHIFT 2Easy

Magnetic Flux & Faraday's Law

A circular disc of radius 20 cm is placed in uniform magnetic field of induction 722\frac{7}{22} Wb/m2^{2} in such a way that its axis makes an angle of 60∘60^\circ with B⃗\vec{B}. The magnetic flux linked with the disc is (cos⁡60∘=0.5)(\cos 60^\circ = 0.5)

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Answer: (b)

  1. Flux Φ=BAcos⁡θ\Phi = BA\cos\theta, where θ\theta is the angle between B⃗\vec{B} and the axis (normal) of the disc.
  2. A=πr2=227(0.2)2=227×0.04A = \pi r^{2} = \frac{22}{7}(0.2)^{2} = \frac{22}{7}\times 0.04 m2^{2}
  3. Φ=722×227×0.04×0.5=0.02\Phi = \frac{7}{22}\times\frac{22}{7}\times 0.04\times 0.5 = 0.02 Wb
  4. So the answer is (b) 0.020.02 Wb.
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