SATHEE: Electromagnetic Induction and Alternating Current 7 Question 24

Electromagnetic Induction and Alternating Current 7 Question 24

####27. A long circular tube of length $10 m$ and radius $0.3 m$ carries a current $I$ along its curved surface as shown. A wire-loop of resistance $0.005 Ω$ and of radius $0.1 m$ is placed inside the tube with its axis coinciding with the axis of the tube.

The current varies as $I = I_{0} \cos 300 t$ where $I_{0}$ is constant. If the magnetic moment of the loop is $N μ_{0} I_{0} \sin (300 t)$ , then $N$ is

(2011)

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Answer:

Correct Answer: 27. 6

Solution:

Take the circular tube as a long solenoid. The wires are closely wound. Magnetic field inside the solenoid is

$B = μ_{0} n i$

Here, $n =$ number of turns per unit length

$∴ n i =$ current per unit length

In the given problem $n i = \frac{I}{L}$

$∴ B = \frac{μ_{0} I}{L}$

Flux passing through the circular coil is

$φ = B S = \frac{μ_{0} I}{L} (π r^{2})$

Induced emf $e = - \frac{d φ}{d t} = - \frac{μ_{0} π r^{2}}{L} \cdot \frac{d I}{d t}$

Induced current, $i = \frac{e}{R} = - \frac{μ_{0} π r^{2}}{L R} \cdot \frac{d I}{d t}$

Magnetic moment, $M = i A = i π r^{2}$

$M = - \frac{μ_{0} π^{2} r^{4}}{L R} \cdot \frac{d I}{d t}$

Given,

$I = I_{0} \cos (300 t)$

$∴ \frac{d I}{d t} = - 300 I_{0} \sin (300 t)$

Substituting in Eq. (i), we get

$\begin{aligned} M = \frac{300 π^{2} r^{4})}{L R} μ_{0} I_{0} \sin (300 t) \\ ∴ N & = \frac{300 π^{2} r^{4}}{L R} \end{aligned}$

Substituting the values, we get

$N = \frac{300 (22 / 7)^{2} (0.1)^{4}}{(10) (0.005)} = 5.926 or N ≃ 6$

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Electromagnetic Induction and Alternating Current 7 Question 24

Answer:

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Electromagnetic Induction and Alternating Current 7 Question 24

Answer:

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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