| ⇦ | 
| ⇨ |
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: - A slab consists of two parallel layers of two different materials of same thickness
- The thermo e.m.f. E in volts of a certain thermocouple is found to vary
- A charge Q is enclosed by a Gaussian spherical surface of radius R
- In a mass spectrometer used for measuring the masses of ions, the ions
- In a P-N junction
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A slab consists of two parallel layers of two different materials of same thickness
- The thermo e.m.f. E in volts of a certain thermocouple is found to vary
- A charge Q is enclosed by a Gaussian spherical surface of radius R
- In a mass spectrometer used for measuring the masses of ions, the ions
- In a P-N junction
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