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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: - A bar magnet with magnetic moment 2.5×10³ JT⁻¹ is rotating in horizontal plane
- Two forces of magnitude 8N and 15N respectively act at a point so as to make the resultant force
- The velocity of sound is V, in air. If the density of air is increased to four times
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Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A bar magnet with magnetic moment 2.5×10³ JT⁻¹ is rotating in horizontal plane
- Two forces of magnitude 8N and 15N respectively act at a point so as to make the resultant force
- The velocity of sound is V, in air. If the density of air is increased to four times
- A parallel plate capacitor of a capacitance 1 pF has seperation between the plates
- A bus is moving with a speed of 10 ms⁻¹ on a straight road
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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