The magnitude of the induced electric field is (RdB/dt)/4
The induced electric field is gotten from
-∫E.dl = dФ/dt where E = induced electric field, dl = path length vector, Ф = magnetic flux through cylindrical region = AB where A = area of magnetic flux = πR² where R = radius of cylindrical region and B = magnetic field.
So, -∫E.dl = dФ/dt
-∫E.dl = dAB/dt
-∫Edlcos0 = AdB/dt (where E.dl = Edlcos0 = Edl since E and dl are parallel to each other.)
So -∫Edl = πR²dB/dt
-E∫dl = πR²dB/dt (∫dl = 2πr since the integral is the circumference of the path)
-E(2πr) = πR²dB/dt (we integrate dl from r = 0 to 2R)
-E2π(2R - 0) = πR²dB/dt
-E4πR= πR²dB/dt
E = πR²dB/dt ÷ 4πR
E = -(RdB/dt)/4
So, the magnitude of the induced electric field is (RdB/dt)/4
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Answer: 55.52 *10^-6 C= 55.52 μC
Explanation: In order to solve this question we have to take into account the following expressions:
potential energy stired in a capacitor is given by:
U=Q^2/(2*C) where Q and C are the charge and capacitance of the capacitor.
then we have:
Q^2= 2*C*U=
C=εo*A/d where A and d are the area and separation of the parallel plates capacitor
Q^2=2*εo*A*U/d=2*8.85*10^-12*1.9*10^-5*11*10^3/(1.2*10^-3)=
=55.52 *10^-6C
Answer:
metal
Explanation:
metal is an excellent conductor for heat and electricity therefore making it perfect to experiment with different temperatures.
Answer:
the velocity of the car is 0.875 m/s
Explanation:
therefore the V of car is 0.875 m
