| ⇦ | 
| ⇨ |
A closely wound solenoid of 2000 turns and area of cross-section 1.5 x 10⁻⁴ m² carries a current of 2.0 A. It suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 x 10⁻² tesla making an angle of 30with the axis of the solenoid. The torque on the solenoid will be:
Options
(a) 3 x 10⁻² N-m
(b) 3 x 10⁻³ N-m
(c) 1.5 x 10⁻³ N-m
(d) 1.5 x 10⁻² N-m
Correct Answer:
1.5 x 10⁻² N-m
Explanation:
Torque on the solenoid is given by ? = MB sin θ
where θ is the angle between the magnetic field and the axis of solenoid. M = niA
? = niA B sin 30⁰
= 2000 x 2 x 1.5 x 10⁻⁴ x 5 x 10⁻² x 1/2
= 1.5 x 10⁻² N – m
Related Questions:
- An electron in a circular orbit of radius 0.05 nm performs 10¹⁶ rev/s. The magnetic
- If the magnetising field on a ferromagnetic material is increased, its permeability
- The periodic waves of intensities I₁ and I₂ pass through a region at the same time
- The ratio of minimum wavelengths of Lyman and Balmer series will be
- The ratio of radii of planets A and B is K₁ and ratio of accelerations
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