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JEE Main 2019 (9 Apr, Shift 2) · Moving Coil Galvanometer, Ammeter & Voltmeter

Moving Coil Galvanometer, Ammeter & Voltmeter question from JEE Main 2019 (9 Apr, Shift 2)

Q20JEE Main 2019 · 9 APR · SHIFT 2Medium

Moving Coil Galvanometer, Ammeter & Voltmeter

A moving coil galvanometer has a coil with 175 turns and area 1 cm2. It uses a torsion band of torsion constant 10–6 N-m/rad. The coil is placed in a maganetic field B parallel to its plane. The coil deflects by 1° for a current of 1 mA. The value of B (in Tesla) is approximately :-

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Answer: (d)

  1. In equilibrium the magnetic torque balances the restoring torque: NIAB=kθNIAB = k\theta.
  2. θ=1∘=π180≈1.745×10−2\theta = 1^\circ = \frac{\pi}{180} \approx 1.745 \times 10^{-2} rad, A=1 cm2=10−4 m2A = 1\text{ cm}^2 = 10^{-4}\text{ m}^2.
  3. B=kθNIA=10−6×1.745×10−2175×10−3×10−4=1.745×10−81.75×10−5≈10−3B = \frac{k\theta}{NIA} = \frac{10^{-6} \times 1.745 \times 10^{-2}}{175 \times 10^{-3} \times 10^{-4}} = \frac{1.745 \times 10^{-8}}{1.75 \times 10^{-5}} \approx 10^{-3} T
  4. So the answer is (d) 10−310^{-3} T.
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