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PhysicsMagnetismMagnetic torque & energyMedium2 minQB

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

Options:

Answer:

Solution:

From energy conservation

$\frac{1}{2} {Iω}^{2} = U_{n} - U_{t}$

$= - MBcos 60 ° - (- MB)$

$\frac{MB}{2} = \frac{1}{2} {Iω}^{2}$

$\frac{20 \times 4}{2} = \frac{1}{2} (08) m^{2}$

$100 = ω^{2}$

$ω = 10 rad$

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Magnetic torque & energy

⏱ 2mℹ️ Source: QB

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