ELECTROMAGNETIC INDUCTION
(Physics)
DPP – 1
SOLUTION
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1.
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ϕ = 8t 2 − 9t + 5, R = 20Ω,
t = 0.25 s
dϕ
e 16t − 9
e = | | = 16t − 9; i = =
dt
R
20
(16 × 0.25) − 9 −5
i(t = 0.25 s) =
=
20
20
−5
i=
× 1000 mA = −250 mA
20
2.
As, the loop is placed in X − Y plane,
⃗B = 3t 3 ĵ + 3t 2 k̂ or
⃗z
dB
= 6t
dt
Now, the induced emf,
e=A
dBz
= πr 2 (6t)
dt
At, t = 2 s, ε = π(1)2 (6 × 2) = 12πV
On comparing with given value, we get n = 12
3.
ϕB (t) = 10t 2 + 20tmwb
∵e=
e=
dϕ
dt
d
(10t 2 + 20t)
dt
e = (20t + 20)mV
at t = 5 seconds
e = 120mV
i=
e 120
=
= 60
R
2
i = 60 mA
APNI KAKSHA
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ELECTROMAGNETIC INDUCTION
(Physics)
Link to View Video Solution:
4.
dϕ
e = − dt = −
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BdA
dt
dr
e = −B × 2πr ( )
dt
22
e = −2 × 2 ×
× (−2 × 10−2 × 10−3 )
7
e = 8π × 10−5 Volt.
5.
e=−
Δϕ
t
=
2NBA
t
e
q = it
R
2NBA
q=
⋅t
R⋅t
2NBA
q=
R
e R
[ = ]
q t
i=
6.
⃗ ⋅A
⃗ = 4 + 12 = 16ωb.
ϕ=B
7.
(A) ACω
8.
e = − dt
(B) Cω
dϕ
(C) Coil I − ACW, Coil 2 − Cw
1
e ∝ dt
1
i ∝ dt
q ∝ dt 0
9.
Induced voltage will be with top as negative and bottom as positive
10.
Here, B = B0 e−τ
t
t
Area of the circular loop, A = πr 2 Flux linked with the loop at any time, t, ϕ = BA = πr 2 B0 e−τ
Emf induced in the loop,
e=−
dϕ
1 t
= πr 2 B0 e−τ
dt
τ
Net heat generated in the loop
∞ e2
= ∫0
=
=
R
π2 r4 B20
τ2 R
dt =
×
−π2 r4 B20
2τ2 R
π2 r4 B20
1
2
(− )
τ
τ2 R
2t
∞
−
∫0 e τ dt
2t
× [e− τ ]
∞
0
× τ(0 − 1) =
π2 r4 B20
2τR
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