Answer:
a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2
b) See the picture
Explanation:
We can use Gauss's law to find the electric field in all the regions:
EA = qen/e0 where qen is the enclosed charge
Remember that the electric field everywhere outside a sphere is:
E(r) = q/(4*pi*eo*r^2) = Kq/r^2
a)
- For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
- For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
- For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2
b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance
Answer:
(a) The current in the wire is 19.89 A
(b) The distance from the wire is 0.159 cm
Explanation:
Given;
magnetic field, B = 2.5 mT
diameter of the wire, d = 1 cm
radius of the wire, r = 0.5 cm = 0.005 m
(a) The current in the wire is calculated as;
![I = \frac{2Br}{\mu_0} \\\\I = \frac{2\times 2.5 \times 10^{-3} \times 0.005 }{4\pi \times 10^{-7} } \\\\I = 19.89 \ A](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B2Br%7D%7B%5Cmu_0%7D%20%5C%5C%5C%5CI%20%3D%20%5Cfrac%7B2%5Ctimes%202.5%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%200.005%20%7D%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%20%7D%20%5C%5C%5C%5CI%20%3D%2019.89%20%5C%20A)
(b) The distance from the wire where the magnetic field is 2.5 mT is calculated as;
![B = \frac{\mu_0 I}{2\pi d} \\\\where;\\\\d \ is \ the \ distance \ from \ the \ wire\\\\d = \frac{\mu_0 I}{2\pi B} = \frac{4 \pi \times 10^{-7} \times 19.89}{2\pi \times 2.5 \times 10^{-3}} = 0.00159 \ m = 0.159 \ cm](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_0%20I%7D%7B2%5Cpi%20d%7D%20%5C%5C%5C%5Cwhere%3B%5C%5C%5C%5Cd%20%5C%20is%20%5C%20the%20%5C%20distance%20%5C%20from%20%5C%20the%20%5C%20%20wire%5C%5C%5C%5Cd%20%3D%20%5Cfrac%7B%5Cmu_0%20I%7D%7B2%5Cpi%20B%7D%20%3D%20%5Cfrac%7B4%20%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%20%5Ctimes%2019.89%7D%7B2%5Cpi%20%5Ctimes%202.5%20%5Ctimes%2010%5E%7B-3%7D%7D%20%20%3D%200.00159%20%5C%20m%20%3D%200.159%20%5C%20cm)
Physical changes: breaking sticks, boiling the water
chemical changes: lighting the fire
(Breaking the wood does not change the substance of the wood , therefore it's a physical change. Water changing between solid, liquid, and gas form is always a physical change.)
Answer: It is done to prevent the necessary compound from solidifying along with the debasements. It expels any insoluble pollutions from the appropriate response (as opposed to separating the predetermined item). With since quite a while ago stemmed channels, the gems kick off inside the progression because the arrangement cools, obstructing the pipe. utilizing a stemless channel keeps this from occurring.
Explanation:
This is probably referring to the process of nuclear fission.