| ⇦ | 
| ⇨ |
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: - The ultimate individual unit of magnetism in any magnet is called
- The potential energy of particle in a force field is U = A/r² – b/r, where A and B
- The 6563 Å line emitted by hydrogen atom in a star is found to be red shifted by 5Å
- The magnetic flux linked with a coil at any instant t is given by ?, find emf
- A body of length 1 m having cross-sectional area 0.75 m² has heat flow through it
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The ultimate individual unit of magnetism in any magnet is called
- The potential energy of particle in a force field is U = A/r² – b/r, where A and B
- The 6563 Å line emitted by hydrogen atom in a star is found to be red shifted by 5Å
- The magnetic flux linked with a coil at any instant t is given by ?, find emf
- A body of length 1 m having cross-sectional area 0.75 m² has heat flow through it
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