SATHEE: magnetism-and-matter Question 3

चुंबकत्व और पदार्थ

Question: Q. 1. A rod of length $L$ , along East West direction is dropped from a height $H$ . If $B$ be the magnetic field due to earth at that place and angle of dip is $θ$ , then what is the magnitude of induced emf across two ends of the rod when the rod reaches the earth.

U] [CBSE SQP, 2016]

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Solution:

Ans.

$\begin{aligned} (1) & ε & = B l v & = B \cos θ \times L \times (2 gH)^{1 / 2} \end{aligned}$

[CBSE Marking Scheme, 2016]

Detailed Answer :

If a rod of length $L$ is dropped from the height $H$ , it will cut the horizontal component of earth’s magnetic field thereby changing the magnetic flux by inducing an emf. When $B$ is the magnetic field and $θ$ is the angle of dip, then horizontal magnetic field will be $B \cos θ$ and velocity of rod dropped will be $\sqrt{2 g H}$ Hence, induced emf will be :

$\begin{aligned} (1) & | e | = B \cos θ \times L \times v & | e | = B \cos θ \times L \times \sqrt{2 g H} \end{aligned}$

AI Q. 2. An electron in an atom revolves around the nucleus in an orbit of radius $r$ with frequency $v$ . Write the expression for the magnetic moment of the electron.

A [Foreign III 2014]

Ans. An electron revolving around the nucleus in an orbit will behave as current loop, so it will possess magnetic momentum,

$M = I A$

where,

$I =$ current flowing through circular loop

Now, $I = \frac{e}{T}$ where, $T =$ time period $= \frac{1}{v}$

$So, \frac{1}{T} = v$

The area of circular loop $A = π r^{2}$

Hence from above, $M = e v (π r^{2})$

[CBSE Marking Scheme 2014]

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चुंबकत्व और पदार्थ

विषयसूची

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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