A circular loop of area 3sq. M and resistance 3ohmis lying in yz plane. The magnetic induction of the location of the loop varies time as B=(2t3-3t2)i+3tj. The magnitude of maximum current in the loop during intervals from t=0 and t=2sec.
Answers
Answer:
okay
0.002 V/m
Given that,
Radius r=2cm
dt
dB
=2T
We know that
Magnetic flux
ϕ=BA
ϕ=B(πr
2
)
Now, induced e. m.f
E=
dt
dϕ
E=
dt
d(Bπr
2
)
E=πr
2
dt
dB
E=3.14×(0.02)
2
×2
E=0.002V
Hence, the induced e. m. f is 0.002 V
Explanation:
hope u have been understood
Given info : A circular loop of area 3 m² and resistance 3Ω lying in yz plane. The magnetic induction of the location of the loop varies time as B = (2t³ - 3t²) i+3t j.
To find : The magnitude of maximum current in the loop during intervals from t = 0 and t = 2 sec is..
solution : circular loop is lying in yz plane so area vector must be perpendicular to it. i.e., A = 3 i
negative of rate of change of magnetic flux = induced potential
⇒V = -dΦ/dt = -d(B.A)/dt
= -d[(2t³ - 3t²)i + 3t j).(3 i) ]/dt
= -d[(6t³ - 9t²)]/dt
= -(18t² - 18t)
= 18t - 18t²
at t = 0, V = 0
at t = 1 , V = 18 × 2 - 18 × 4 = -36
so potential difference = (0 - (-36)) = 36 volts
from Ohm's law, I = V/R
= 36 volts/3 Ω = 12 A
Therefore the current in the loop during intervals t = 0 to t = 2 sec, is 12 A.