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:
brainly.com/question/15730392
The first thing you should know is that the work is defined as:
W = F * d
Where
F = force
d = displacement
We have then
(a) the block
F = (0.2) * (100) = 20
d = 100
W = (20) * (100) = 2000 ft.lbf
(b) the man as the system.
F = (0.2) * (100 + 180) = 56
d = 100
W = (56) * (100) = 5600 ft.lbf
answer:
(a) 2000 ft.lbf
(b) 5600 ft.lbf
Stars are made of very hot gas. This gas is mostly hydrogen and helium, which are the two lightest elements. Stars shine by burning hydrogen into helium in their cores, and later in their lives create heavier elements.
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