SATHEE: Magnetics 6 Question 26

Magnetics 6 Question 26

29. A ring of radius $R$ having uniformly distributed charge $Q$ is mounted on a rod suspended by two identical strings. The tension in strings in equilibrium is $T_{0}$ . Now, a vertical magnetic field is switched on and ring is rotated at constant angular velocity $ω$ . Find the maximum $ω$ with which the ring can be rotated if the strings can withstand a maximum tension of $3 T_{0} / 2$ .

$(2003, 4$ M)

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Solution:

In equilibrium, $2 T_{0} = m g$ or $T_{0} = \frac{m g}{2}$

Magnetic moment, $M = i A = \frac{ω}{2 π} Q (π R^{2})$

$τ = M B \sin 90^{\circ} = \frac{ω B Q R^{2}}{2}$

Let $T_{1}$ and $T_{2}$ be the tensions in the two strings when magnetic field is switched on $(T_{1} > T_{2})$ .

For translational equilibrium,

$T_{1} + T_{2} = m g$

For rotational equilibrium

$(T_{1} - T_{2}) \frac{D}{2} = τ = \frac{ω B Q R^{2}}{2} or T_{1} - T_{2} = \frac{ω B Q R^{2}}{2}$

Solving Eqs. (ii) and (iii), we have

$T_{1} = \frac{m g}{2} + \frac{ω B Q R^{2}}{2 D}$

As $T_{1} > T_{2}$ and maximum values of $T_{1}$ can be $\frac{3 T_{0}}{2}$ , we have

$\begin{aligned} \frac{3 T_{0}}{2} & = T_{0} + \frac{ω_{max} B Q R^{2}}{2 D} \frac{m g}{2} = T_{0} \\ ∴ ω_{max} & = \frac{D T_{0}}{B Q R^{2}} \end{aligned}$

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Magnetics 6 Question 26

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Magnetics 6 Question 26

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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