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:
value of fs,max under the circumstances is
f = 42.56 N
Explanation:
The formula is µ = f / N, where µ is the coefficient of friction, f is the amount of force that resists motion, and N is the normal force. Normal force is the force at which one surface is being pushed into another.
So f = µN
f = 0.380 * 112
f = 42.56 N
Answer:48.52 kJ
Explanation:
Given
Resistance
temperature increases from 
Voltage=50 V
Heat given(H)
Where V=voltage applied
t=time
R=Resistance
Heat absorbed by water is
where
m=mass of water
c=specific heat of water
=change in temperature
Therefore 90-41.48=48.52 kJ is not absorbed by water and leaves the system into the surroundings.
