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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: - which of the following can not be emitted by radioactive substances during their decay?
- The current through an inductor changes from 3 A to 2 A is 1 m/s. If induced e.m.f.
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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
- which of the following can not be emitted by radioactive substances during their decay?
- The current through an inductor changes from 3 A to 2 A is 1 m/s. If induced e.m.f.
- A radiation of energy ‘E’ falls normally on a perfectly reflecting surface
- A system consists of three masses m₁, m₂ and m₃ connected by a string passing
- The ionisation energy of an electron in the ground state of helium atom is 24.6 eV.
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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