Answer:
E = 10t^2e^-10t Joules
Explanation:
Given that the current through a 0.2-H inductor is i(t) = 10te–5t A.
The energy E stored in the inductor can be expressed as
E = 1/2Ll^2
Substitutes the inductor L and the current I into the formula
E = 1/2 × 0.2 × ( 10te^-5t )^2
E = 0.1 × 100t^2e^-10t
E = 10t^2e^-10t Joules
Therefore, the energy stored in the inductor is 10t^2e^-10t Joules
Answer:
2.12/R mW
Explanation:
The electrical power, P generated by the rod is
P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?
So, the induced emf on the conductor is
E = BLv
= 0.575 T × 0.2 m × 0.4 m/s
= 0.046 V
= 46 mV
The electrical power, P generated by the rod is
P = B²L²v²/R
= B²L²v²/R
So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²
= 0.002116/R W
= 2.12/R mW
Explanation:
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