Answer: left and applied force
Explanation:
Answer:
(a) V1 = 8990.00 V
V2 = 8960.13 V
Explanation:
Parameters given:
q =3 mC
k = 8.99 * 10⁹ Nm²/C²
x1 = 3 m
x2 = 3.01 m
Electric potential is given as:
V = kq/r
Where
k = Coulombs constant
q = charge
r = distance
Potential at x1 is:
V1 = (8.99 * 10⁹ * 0.000003)/(3)
V1 = 8990.00V
Potential at x2 is:
V2 = (8.99 * 10⁹ * 0.003)/(3.01)
V2 = 8960.13 V
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Answer:
The average induced emf in the loop is 0.20 V
Explanation:
Given:
Radius of loop
m
Magnetic field
T
Change in time
sec
According to the faraday's law,
Induced emf is given by

Where
magnetic flux
( here
)
Where 
We neglect minus sign because it's shows lenz law


V
Therefore, the average induced emf in the loop is 0.20 V