Magnetic flux in a circuit containing a coil of resistance 2ohm changes from 2 wb to10wb in 0.2 sec the charge passed through the coil in this time is
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Answered by
52
induced emf in coil (E) = change in Φₘ / t
= (10 - 2) / 0.2
= 40
now , E = I . r
I = 40 / 2
I = 20 Amp
I.t = q
=> q = 20 x 0.2 = 4 C
= (10 - 2) / 0.2
= 40
now , E = I . r
I = 40 / 2
I = 20 Amp
I.t = q
=> q = 20 x 0.2 = 4 C
Answered by
7
Dear Student,
◆ Answer -
Q = 4 C
● Explanation -
# Given -
Φ1 = 2 Wb
Φ2 = 10 Wb
R = 2 ohm
t = 0.2 s
# Solution -
EMF induced in the coil is -
V = ∆Φ/dt
V = (Φ2-Φ1)/dt
V = (10-2)/0.2
V = 40 V
Charge passed through the coil is -
Q = I.t
Q = (V/R).t
Q = (40/2) × 0.2
Q = 4 C
Hence, charge passed through the coil is 4 C.
Thanks dear. Hope this helps you...
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