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A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60⁰. The torque needed to maintain the needle in this position will be:
Options
(a) 2 √3 J
(b) 3 J
(c) √3 J
(d) 3/2 J
Correct Answer:
3 J
Explanation:
According to work energy theorem
W = U final – U initial = MB (cos 0 – cos 60⁰)
W = MB/2 = √3J …(i)
? = M⃗ x B⃗ = MB sin 60⁰ = (MB √3 / 2) …(ii)
From eq (i) and (ii)
? = 2 √3 x √3 / 2 = 3 J
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Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A certain current on passing through a galvanometer produces a deflection
- A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV.
- To what temperature should the hydrogen at room temperature (27°C) be heated at constant
- A capacitor is charged to 200 volt. It has a charge of 0.1 coulomb.
- Two balls of mass m₁ and m₂ are separated from each other by powder charge
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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