| ⇦ | 
| ⇨ |
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 electric potential at the surface of an atomic nucleus (z=50) of radius
- A galvanometer coil has a resistance of 15 ohm and gives full scale deflection
- When one of the slits of Young’s experiment is covered with a transport sheet
- Magnetic susceptibility of a material is given as
- The charge flowing through a resistance R varies with time t as Q=at-bt²,
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The electric potential at the surface of an atomic nucleus (z=50) of radius
- A galvanometer coil has a resistance of 15 ohm and gives full scale deflection
- When one of the slits of Young’s experiment is covered with a transport sheet
- Magnetic susceptibility of a material is given as
- The charge flowing through a resistance R varies with time t as Q=at-bt²,
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