SATHEE: Electromagnetic Induction and Alternating Current 3 Question 8

Electromagnetic Induction and Alternating Current 3 Question 8

####9. A circular wire loop of radius $R$ is placed in the $x - y$ plane centred at the origin $O$ . A square loop of side $a (a ≪ R)$ having two turns is placed with its centre at $z = \sqrt{3} R$ along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of $45^{\circ}$ with respect to the $Z$ -axis.

If the mutual inductance between the loops is given by

$\frac{μ_{0} a^{2}}{2^{p / 2} R}$ , then the value of $p$ is

(2012)

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Answer:

Correct Answer: 9. 7

Solution:

If $I$ current flows through the circular loop, then magnetic flux at the location of square loop is

$B = \frac{μ_{0} I R^{2}}{2 {(R^{2} + Z^{2})}^{3 / 2}}$

Substituting the value of $Z (= \sqrt{3} R)$ ,

we have

$B = \frac{μ_{0} I}{16 R}$

Now, total flux through the square loop is

$φ_{T} = N B S \cos θ = (2) \frac{μ_{0} T}{16 R} a^{2} \cos 45^{\circ}$

Mutual inductance,

$\begin{aligned} M = \frac{φ_{T}}{I} = \frac{μ_{0} a^{2}}{2^{7 / 2} R} \\ ∴ p & = 7 \end{aligned}$

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Electromagnetic Induction and Alternating Current 3 Question 8

Answer:

Solution:

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Electromagnetic Induction and Alternating Current 3 Question 8

Answer:

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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