Answer:
I = 4.642 Ampere.
x = 2.145 cm
Explanation:
a) As we know, the magnetic field on the axis of the loop is given as
![B = \frac{ \mu NIa^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} } }](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%09%5Cmu%20NIa%5E%7B2%7D%20%7D%7B2%28x%5E%7B2%7D%20%2B%20a%5E%7B2%7D%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%20%7D)
where a = radius of loop
x = point on the axis of loop
N = No of turns of coil
Current in the loop for which the magnetic field at the center is 0.0750 Tesla is given as x = 0
Therefore, the above equation can be rewritten as
![B_{x} = \frac{ \mu NI}{2a}](https://tex.z-dn.net/?f=B_%7Bx%7D%20%20%3D%20%5Cfrac%7B%09%5Cmu%20NI%7D%7B2a%7D)
I = ![\frac{2aB}{μN}](https://tex.z-dn.net/?f=%5Cfrac%7B2aB%7D%7B%CE%BCN%7D)
by putting values
I =![\frac{2 X 0.0750 T X 0.028 m}{4\pi X 10^{-7} T.\frac{m}{A} X 720}](https://tex.z-dn.net/?f=%5Cfrac%7B2%20X%200.0750%20T%20X%200.028%20m%7D%7B4%5Cpi%20X%2010%5E%7B-7%7D%20T.%5Cfrac%7Bm%7D%7BA%7D%20%20X%20720%7D)
Current in the loop for which the magnetic field at the center is 0.0750 Tesla = I = 4.642 Ampere
b)
Now for part b the magnetic field at a distance x from the center is given as
![B = \frac{ \mu N I a^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} } }](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%09%5Cmu%20N%20I%20a%5E%7B2%7D%20%7D%7B2%28x%5E%7B2%7D%20%2B%20a%5E%7B2%7D%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%7D)
multiply and divide by a on both sides we get
![B = \frac{ \mu NI}{2a} X \frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%09%5Cmu%20NI%7D%7B2a%7D%20X%20%5Cfrac%7Ba%5E%7B3%7D%20%7D%7B%28x%5E%7B2%7D%20%2B%20a%5E%7B2%7D%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%20%7D)
As we know, according to Biot sivorts law,the Magnetic Field at the Center of a circular loop is given as
( Magnetic field at center) = ![B_{c}](https://tex.z-dn.net/?f=B_%7Bc%7D)
So we got magnetic field at any point x as
=
X ![\frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }](https://tex.z-dn.net/?f=%5Cfrac%7Ba%5E%7B3%7D%20%7D%7B%28x%5E%7B2%7D%20%2B%20a%5E%7B2%7D%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%20%7D)
For magnetic field at x is half of the B at center
=
from the above two equations
= ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
![(x^{2} + a^{2})^{3} = 4a^{6}](https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2B%20a%5E%7B2%7D%29%5E%7B3%7D%20%3D%204a%5E%7B6%7D)
Putting a = 2.80 cm
We have x = 2.145 cm Ans
μ
If the train stops then the math is gonna be c4
Given:
Temperature of water,
=
=273 +(-6) =267 K
Temperature surrounding refrigerator,
=
=273 + 21 =294 K
Specific heat given for water,
= 4.19 KJ/kg/K
Specific heat given for ice,
= 2.1 KJ/kg/K
Latent heat of fusion,
= 335KJ/kg
Solution:
Coefficient of Performance (COP) for refrigerator is given by:
Max
= ![\frac{T_{2}}{T_{2} - T_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B2%7D%20-%20T_%7B1%7D%7D)
=
= 9.89
Coefficient of Performance (COP) for heat pump is given by:
Max
= ![\frac{T_{1}}{T_{2} - T_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%20-%20T_%7B1%7D%7D)
= 10.89