Physics Ncert Solutions Class 12th Electromagnetic Induction Class 12th Physics

A planar loop of wire rotates in a uniform magnetic field. Initially, at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mn>2.5</mn> <mtext> </mtext> <mi mathvariant="normal">s</mi> </math> </span> and <span class="mathml" contenteditable="false"> <math> <mn>5.0</mn> <mtext> </mtext> <mi mathvariant="normal">s</mi> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mn>5.0</mn> <mtext> </mtext> <mi mathvariant="normal">s</mi> </math> </span>and <span class="mathml" contenteditable="false"> <math> <mn>10.0</mn> <mtext> </mtext> <mi mathvariant="normal">s</mi> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mn>2.5</mn> <mtext> </mtext> <mi mathvariant="normal">s</mi> </math> </span>and <span class="mathml" contenteditable="false"> <math> <mn>7.5</mn> <mtext> </mtext> <mi mathvariant="normal">s</mi> </math> </span></p>

Option 4 - <p>5.0 and <span class="mathml" contenteditable="false"> <math> <mn>7.5</mn> <mtext> </mtext> <mi mathvariant="normal">s</mi> </math> </span><math><mtext></mtext></math></p>

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Answered by

Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 1

Detailed Solution:

At any time $t$

$? = B A c o s ? \frac{π t}{5} (? ω = \frac{2 π}{T} = \frac{2 π}{10} = \frac{π}{5})$

$\frac{d ?}{d t} = - \frac{π}{5} B A s i n ? (\frac{π t}{5})$

$e = \frac{π}{5} B A s i n ? (\frac{π t}{5})$

e will be maximum when $\frac{π t}{5}$ is $\frac{π}{2}$ .

$t = 2.5 s e c$
e will be minimum when $\frac{π t}{5}$ is $π, 0$

$t = 0,5 s e c$

So, answer will be (1).

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A rod AB moves with a uniform velocity v in a uniform magnetic field as shown in figure then :-

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An aeroplane, with its wings spread 10m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth’s field at that part is 2.5 × 10^-⁴Wb/m² and the angle of dip is 60°. The emf induced between the tips of the plane wings will be…………..

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B_v = B sin 60°

-> $B_{v} = 2.5 \times 1 0^{- 4} \times \frac{\sqrt[]{3}}{2}$

$E m f = B_{v} \times v \times l = 2.5 \times 1 0^{- 4} \times \frac{\sqrt[]{3}}{2} \times 180 \times \frac{5}{18} \times 1 = 108.25 \times 1 0^{- 3} v o l t s$

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Figure shows a circuit that contains four identical resistors with resistance R = $2.0 Ω,$  two identical inductors with inductance L = 2.0mH and an ideal battery with emf E = 9V. The current ‘i’ just after switch ‘S’ is closed will be:

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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