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JEE Main 20204 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
4 September
Shift
Evening Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2020 · 4 September · Evening Shift · Physics · Magnetic Effects of Current

A circular coil has moment of inertia 0 . 8kg m ^ 2 around any...

A circular coil has moment of inertia 0.8kgm20 . 8kg{m}^{2} around any diameter and is carrying current to produce a magnetic moment of 20Am220{Am}^{2}. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4T4T is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60o{60}^{o} will be :

Options

  1. A10rads−110rad{s}^{- 1}
  2. B10πrads−110\pi\text{rad}{s}^{- 1}
  3. C20πrads−120\pi\text{rad}{s}^{- 1}
  4. D20rads−120rad{s}^{- 1}

Correct answer

A. 10rads−110rad{s}^{- 1}

Step-by-step solution

From energy conservation

12Iω2=Un−Ut\dfrac{1}{2}{I \omega}^{2} = {U}_{n} - {U}_{t}

=−MBcos60°−(−MB)= - MBcos60 ° - ( - MB )

MB2=12Iω2\dfrac{MB}{2} = \dfrac{1}{2}{I \omega}^{2}

20×42=12(08)m2\dfrac{20 \times 4}{2} = \dfrac{1}{2} ( 08 ) {m}^{2}

100=ω2100 = {\omega}^{2}

ω=10rad\omega = 10rad

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