Electromagnetic Induction

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New answer posted

3 weeks ago

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A
alok kumar singh

Contributor-Level 10

Bv = B sin 60°

-> B v = 2 . 5 * 1 0 4 * 3 2

E m f = B v * v * l = 2 . 5 * 1 0 4 * 3 2 * 1 8 0 * 5 1 8 * 1 = 1 0 8 . 2 5 * 1 0 3 v o l t s        

New answer posted

4 weeks ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

M = φ? /I? = (B? A? )/I? = [ (μ? I? /2R? )πR? ²]/I?
[Diagram of two concentric coils]
M = (μ? πR? ²)/ (2R? )
M ∝ R? ²/R?

New question posted

a month ago

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New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

(A) The magnet's entry    

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

R = 2 Ω

L = 2 mH

E = 9V

i = ε 2 R = 9 v 4 Ω = 2 . 2 5 A

Just after the switch 'S' is closed, the inductor acts as open circuit.

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

U B = 1 2 L l 2 . . . . . . . . . . . . ( i )

PR = l2R

=> 640 = (8)2 R

R = 1 0 Ω

from (i) 64 = 1 2 * L * ( 8 ) 2

L = 2H

l = L R = 2 1 0 = 0 . 2 s e c o n d        

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

At any time  t

? = B A c o s ? π t 5 ? ω = 2 π T = 2 π 10 = π 5

d ? d t = - π 5 B A s i n ? π t 5

e = π 5 B A s i n ? π t 5

e will be maximum when π t 5  is π 2 .

t = 2.5 s e c
e will be minimum when π t 5 is π , 0

t = 0,5 s e c

So, answer will be (1).

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Assuming current 'i' In outer loop magnetic field at centre

= 4 * μ 0 4 π i L / 2 * [ 2 s i n 4 5 ° )

= 2 2 μ 0 i π L

M = F m α t h r o u g h i n n e r l o o p i

= 2 2 μ 0 i l 2 π L = 2 2 μ 0 l 2 π L

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

This phenomenon works on resonance in AC circuit.

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

? = B ? A ? = ( 3 i ˆ + 4 k ˆ ) ( 25 i ˆ + 25 k ˆ )

? = ( 3 * 25 ) + ( 4 * 25 ) = 175   weber  

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