⇦ | ![]() | ⇨ |
A bar magnet having a magnetic moment of 2 x 10⁴JT⁻¹ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 x 10⁻⁴T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60⁰ from the field is
Options
(a) 12 J
(b) 6 J
(c) 2 J
(d) 0.6 J
Correct Answer:
6 J
Explanation:
Work done
= MB (cos θ₁ – cos θ₂)
= MB (cos 0⁰ – cos 60⁰)
= MB (1 – 1/2) = 2 x 10⁴ x 6 x 10⁻⁴ /2 = 6 J
Related Questions:
- Acceleration of a particle, executing SHM, at its mean position is
- A boat is sent across a river with a velocity of 8 km/h.If the resultant velocity
- In a P-N junction
- A current of 2A flows through a 2Ω resistor when connected across a battery
- In a laser tube, all the photons
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