Answer:
Explanation:
Given that,
B(t) = B0 cos(ωt) • k
Radius r = a
Inner radius r' = a/2 and resistance R.
Current in the loop as a function of time I(t) =?
Magnetic flux is given as
Φ = BA
And the Area is given as
A = πr², where r = a/2
A = πa²/4
Then,
Φ = ¼ Bπa²
Φ(t) = ¼πa²Bo•Cos(ωt)
Then, the EMF is given as
ε(t) = -dΦ/dt
ε(t) = -¼πa²Bo • -ωSin(ωt)
ε(t) = ¼ωπa²Bo•Sin(ωt)
From ohms law,
ε = iR
Then, i = ε/R
I(t) = ¼ωπa²Bo•Sin(ωt) /R
This is the current induced in the loop.
Check attachment for better understanding
A fundamental force , is your answer .
Answer:
unchanged
Explanation:
Let the voltage of the battery be V
Inductance L1 = 5 mH
Inductance L2 = 10 mH
consider resistance R of the circuit (wire, battery).
V = I R + L dI/dt
where, I is the current in the circuit and t is the time.
After few seconds of connection being made, the factor dI/dt is negligible. There is no change in the current flowing through the circuit. when inductor was just attached in the circuit, a current