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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Decomposition reactions are said to be those reactions in which a reactants breakdown into two or more products. The general reaction for decomposition reactions is as follow,
ABC → A + B + C
Specific Examples are as,
Water → Hydrogen + Oxygen
2 H₂O → 2 H₂ + O₂
Calcium carbonate → Calcium oxide + Carbon dioxide
CaCO₃ → CaO + CO₂
While, Synthetic reactions are said to be those reactions in which two or more reactants combine to form two or more products. The general reaction for synthetic reactions is as follow,
A + B + C → ABC
Specific Examples are as,
Iron + Oxygen → Iron Oxide
2 Fe + 3 O₂ → 2 Fe₂O₃
Sodium + Chlorine → Sodium chloride
2 Na + Cl₂ → 2 NaCl
Sulfur + Oxygen → Sulfur dioxide
S + O₂ → SO₂
Potassium + Chlorine → Potassium chloride
2 K + Cl₂ → 2 KCl
Answer:
40g
Explanation:
Solubility of Copper sulfate at 90°=60g
Solubility of potassium bromide at 90°=100g
100g-60g=40g
Answer:
Explanation:
The given time is 1 / 4 of the time period
So Time period of oscillation.
= 4 x .4 =1.6 s
When the block reaches back its original position when it came in contact with the spring for the first time , the block and the spring will have maximum
velocity. After that spring starts unstretching , reducing its speed , so block loses contact as its velocity is not reduced .
So required velocity is the maximum velocity of the block while remaining in contact with the spring.
v ( max ) = w A = 1.32 m /s.
