Answer:
Current in outer circle will be 15.826 A
Explanation:
We have given number of turns in inner coil ![N_I=170](https://tex.z-dn.net/?f=N_I%3D170)
Radius of inner circle ![r_i=0.0095m](https://tex.z-dn.net/?f=r_i%3D0.0095m)
Current in the inner circle ![I_i=8.9A](https://tex.z-dn.net/?f=I_i%3D8.9A)
Number of turns in outer circle ![N_o=150](https://tex.z-dn.net/?f=N_o%3D150)
Radius of outer circle ![r_o=0.015m](https://tex.z-dn.net/?f=r_o%3D0.015m)
We have to find the current in outer circle so that net magnetic field will zero
For net magnetic field current must be in opposite direction as in inner circle
We know that magnetic field is given due to circular coil is given by
![B=\frac{N\mu _0I}{2r}](https://tex.z-dn.net/?f=B%3D%5Cfrac%7BN%5Cmu%20_0I%7D%7B2r%7D)
For net magnetic field zero
![\frac{N_I\mu _0I_I}{2r_I}=\frac{N_O\mu _0I_0}{2r_O}](https://tex.z-dn.net/?f=%5Cfrac%7BN_I%5Cmu%20_0I_I%7D%7B2r_I%7D%3D%5Cfrac%7BN_O%5Cmu%20_0I_0%7D%7B2r_O%7D)
So ![\frac{170\times \mu _0\times 8.9}{2\times 0.0095}=\frac{150\times \mu _0I_O}{2\times 0.015}](https://tex.z-dn.net/?f=%5Cfrac%7B170%5Ctimes%20%5Cmu%20_0%5Ctimes%208.9%7D%7B2%5Ctimes%200.0095%7D%3D%5Cfrac%7B150%5Ctimes%20%5Cmu%20_0I_O%7D%7B2%5Ctimes%200.015%7D)
![I_O=15.92A](https://tex.z-dn.net/?f=I_O%3D15.92A)
Answer:
"h" signifies Planck's constant
Explanation:
In the equation energy E = h X v
The "h" there signifies Planck's constant
Planck's constant is a value, that shows the rate at which the energy of a photon increases/decreases, as the frequency of its electromagnetic wave changes.
It was named after Max Planck who discovered this unique relationship between the energy of a light wave and its frequency.
Planck's constant, "h" is usually expressed in Joules second
Planck's constant = ![6.62607015 \times 10^{-34} J.s](https://tex.z-dn.net/?f=6.62607015%20%5Ctimes%2010%5E%7B-34%7D%20%20J.s)
Answer:
![H=41.3kJmol^{-1}](https://tex.z-dn.net/?f=H%3D41.3kJmol%5E%7B-1%7D)
Explanation:
The equation relating the the enthalphy, pressure and temperature is expressed as
![ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%20%29%3D%5Cfrac%7BH%7D%7BR%7D%28%5Cfrac%7B1%7D%7BT_%7B1%7D%7D-%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20%29%20%5C%5C)
Where P is the pressure, H is the enthalphy, and T is the temperature.
since the given values are
![T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}](https://tex.z-dn.net/?f=T_%7B1%7D%3D293k%2C%20%5C%5CT_%7B2%7D%3D336.5k%5C%5CP_%7B1%7D%3D5.95kPA%5C%5Cp_%7B2%7D%3D53.3kPA%5C%5Cand%20R%3D8.314J.K%5E%7B-1%7Dmol_%7B-1%7D)
if we insert values, we arrive at
![ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B53.3%7D%7B5.95%7D%20%29%3D%5Cfrac%7BH%7D%7B8.314%7D%28%5Cfrac%7B1%7D%7B293%7D-%5Cfrac%7B1%7D%7B336.5%7D%20%29%5C%5C2.19%3D%5Cfrac%7BH%7D%7B8.314%7D%280.00044%29%5C%5CH%3D%282.19%2A8.314%29%2F0.00044%5C%5CH%3D41%2C268.8Jmol%5E%7B-1%7D%5C%5CH%3D41.3kJmol%5E%7B-1%7D)
I think it was by boat and coming from over seas