| ⇦ | 
| ⇨ |
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: - From a disc of radius R, a concentric circular portion of radius r is cut-out so
- Two bodies of mass 10 kg and 5 kg moving in concentric orbits of radius
- The current flowing through a lamp marked as 50 W and 250 V is
- The radius of the first orbit of the hydrogen atom is a₀. The radius of the second orbit
- The equation of state for 5 g of oxygen at a pressure P and temperature T,
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- From a disc of radius R, a concentric circular portion of radius r is cut-out so
- Two bodies of mass 10 kg and 5 kg moving in concentric orbits of radius
- The current flowing through a lamp marked as 50 W and 250 V is
- The radius of the first orbit of the hydrogen atom is a₀. The radius of the second orbit
- The equation of state for 5 g of oxygen at a pressure P and temperature T,
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