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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
Related Questions: - A particle executing simple harmonic motion of amplitude 5 cm has maximum
- A small signal voltage V(t)=V₀ sin ωt is applied across an ideal capacitor C
- A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance
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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 particle executing simple harmonic motion of amplitude 5 cm has maximum
- A small signal voltage V(t)=V₀ sin ωt is applied across an ideal capacitor C
- A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance
- In Young’s double slit experiment, the ratio of maximum and minimum intensities
- A carnot’s engine operates with source at 127° C and sink at 27°C
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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