A circular coil has moment of inertia 0.8 kg m² around any diameter and is carrying current to produce a magnetic moment of 20Am². The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4 T is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60° will be:

Option 1 - <p>10πrads⁻¹<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>20πrads⁻¹<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>20rads⁻¹<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>10rads⁻¹<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
4 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago
Correct Option - 3
Detailed Solution:

I = 0.8 kgM², |μ? | = 20Am²
U? = -μ? ⋅ B? = 0
Uf = -μBcos (30°) = -20 x 4 x √3/2
U? – Uf = 40√3 = ½ I ω² = 0.4ω²
ω² = 100√3; ω = 10 (3¹/? )

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Physics System of Particles and Rotational Motion 2025

Physics System of Particles and Rotational Motion 2025

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