Electromagnetic Induction

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New answer posted

2 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

i = E R = 1 R d ? d t = a 2 R d B d t

i = ( 7.5 ) 2 * 10 - 4 * ? * 4 * 10 - 6 1.23 * 10 - 8 * 0.3 * 0.032

= 0.61 A

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

For a rotating loop, the induced emf is ε = ε? sin (ωt), where the peak emf ε? = NABω.
Here, Area A = πab. The number of turns N = 1.
So, ε? = B (πab)ω
The average power loss due to Joule heating in a resistor R is given by:
P_avg = (ε_rms)² / R
Where ε_rms = ε? /√2
P_avg = (ε? ²/2) / R = (B (πab)ω)² / (2R) = π²a²b²B²ω² / (2R)

New answer posted

2 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

Flux as a function of time Φ = B.A = ABcos (ωt) emf induced,
e = -dΦ/dt = ABωsin (ωt)
Maximum value of emf = ABω = πR²Bω
= 3.14*0.1*0.1*3*10? * (0/0.2) = 15

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Vq - 2*1 + 30 + 50*10? ³*10² = Vp
⇒ Vp-Vq = 33.0

New question posted

2 months ago

0 Follower 3 Views

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

As magnetic field lines always form a closed loop, hence every magnetic field line creating magnetic flux in the inner region must be passing through the outer region. Since flux in two regions are in opposite direction,

? i = - ? 0

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

ε = - d φ d t = - A d B d t

= ( 16 * 4 - 4 * 2 ) ( 1000 - 500 ) 5 * 10 - 4 * 10 - 4

= 56 * 500 5 * 10 - 8 = 56 * 10 - 6 V

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

  I = I 0 t - I 0 t 2

? = B A

? = μ 0 n I A

V R = d ? d t = - μ 0 n A l 0 ( 1 - 2 t )

V R = 0 at   t = 1 2 and I R = V R   Resistance of loop  

 

New answer posted

2 months ago

0 Follower 5 Views

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).

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