| ⇦ | 
| ⇨ |
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: - Eddy currents are used in
- A wire of length 1 m is moving at a speed of 2 ms⁻¹ perpendicular to its length
- A nucleus of mass number 220, initially at rest, emits an α-particle. If the Q-value
- Light emitted during the de excitation of electron from n=3 to n=2, when incident
- A heavy stone hanging from a massless string of length 15 m is projected
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Eddy currents are used in
- A wire of length 1 m is moving at a speed of 2 ms⁻¹ perpendicular to its length
- A nucleus of mass number 220, initially at rest, emits an α-particle. If the Q-value
- Light emitted during the de excitation of electron from n=3 to n=2, when incident
- A heavy stone hanging from a massless string of length 15 m is projected
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