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
Learn more about induced electric field here:
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Answer:
C. Sources of electrical current!
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Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects, usually denser than water, to float and slide on a water surface.
