⇦ | ⇨ |
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: - In L-C-R series circuit, an alternating emf e and current i are given by the equations
- In case of linearly polarised light, the magnitude of the electric field vector
- An electron of a stationary hydrogen atom passes from the fifth energy level to
- Weight of a body of mass m decreases by 1% when it is raised to height h
- In Young’s double slit experiment, the intensity of light coming from the first slit
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In L-C-R series circuit, an alternating emf e and current i are given by the equations
- In case of linearly polarised light, the magnitude of the electric field vector
- An electron of a stationary hydrogen atom passes from the fifth energy level to
- Weight of a body of mass m decreases by 1% when it is raised to height h
- In Young’s double slit experiment, the intensity of light coming from the first slit
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply