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 solution would
be like this for this specific problem:
<span>
F=−</span>k∗x∗<span>q∗</span>Q<span>/(</span>+)<span>F−≈</span><span><span><span>k∗x∗<span>q∗</span>Q</span><span>/R3</span></span>[(</span>1−<span><span>3/2</span><span><span>*x2</span><span>/R3</span></span>]
</span><span>F=−</span><span><span>k∗x∗<span>q∗/</span>Q</span><span>R<span>3
</span></span></span><span>F=</span><span>ma
</span>−<span><span><span><span>k∗<span>q∗</span>Q</span><span>/R3</span></span>*</span>x</span>=<span>ma
</span>−k∗x=m∗<span>a
a</span>==<span><span><span>ω2</span>x
</span>ω</span><span>=(</span>k/<span>m<span>)<span><span>1/</span><span>2
</span></span></span></span>ω<span>=(</span><span>kqQ</span>/<span><span>R3</span><span>)<span><span>1/</span>2
</span></span></span>
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Answer:
D. a solid, liquid, or gaseous compound that is found in living organisms
Explanation:
Answer:
<u>Acceleration</u> is the rate of change of velocity of an object.
Formula: Had to take a picture of the formula
Answer:
0.366×10^{-3} / s
Explanation:
θ = θmax e^{-bt/2m}
Given that
θ = 5.50°
θmax = 15.0°
So that we have
ln (θ / θmax) = -bt /2m
= - ln(5.50°/ 15.0°) / 1000s = b /2m
= b / 2m = 0.366×10^{-3} / s
