Answer:
![I = 50.78 A](https://tex.z-dn.net/?f=I%20%3D%2050.78%20A)
Explanation:
Our Given Parameters include:
Average radius (r) for each parallel coil = 12.5 cm = 12.5 × 10⁻² m
The lower coil turns = 20 turns
The lower coil constant current = 1.30 A
The upper is suspended 0.314 cm above the lower coil
That indicates the distance = 0.314 cm = 0.314 × 10⁻² m
Current for the upper coil = ???(unknown)
Force (F) = 3.30 N
If we take each coil into cognizance as a long parallel straight wire; then the length of the lower coil can be calculated as:
![L__L}=N__L}(2 \pi r)](https://tex.z-dn.net/?f=L__L%7D%3DN__L%7D%282%20%5Cpi%20r%29)
where:
= number of turns of the lower coil = 20 turns
r = average radius in the lower coil = 12.5 × 10⁻² m
Substituting our values; we have:
![L__L}=20*(2 \pi (12.5*10^{-2}m)](https://tex.z-dn.net/?f=L__L%7D%3D20%2A%282%20%5Cpi%20%2812.5%2A10%5E%7B-2%7Dm%29)
![L__L}=15.70m](https://tex.z-dn.net/?f=L__L%7D%3D15.70m)
From our parameters above:
= constant current in the lower coil = 4.0 A
But the magnitude of the magnetic force (F) = 1 LB
Then the force on the lower coil in regard to the upper coil can be :
F = ![I__L}L__L}[\frac{u__0I__upper}{2 \pi d}]](https://tex.z-dn.net/?f=I__L%7DL__L%7D%5B%5Cfrac%7Bu__0I__upper%7D%7B2%20%5Cpi%20d%7D%5D)
Making the varied current in the upper coil the subject of the formula; we have:
![I_{upper}= \frac{2 \pi d F}{U_oI__L}L__L}](https://tex.z-dn.net/?f=I_%7Bupper%7D%3D%20%5Cfrac%7B2%20%5Cpi%20d%20F%7D%7BU_oI__L%7DL__L%7D)
where : ![U__0}= 4 \pi *10^{-7}\frac{T.m}{A}](https://tex.z-dn.net/?f=U__0%7D%3D%204%20%5Cpi%20%2A10%5E%7B-7%7D%5Cfrac%7BT.m%7D%7BA%7D)
Then:
![I_{upper}= \frac{2 \pi (0.314*10^{-2}(3.30)}{ (4 \pi *10^{-7}\frac{T.m}{A})(1.30A)(15.70m)}](https://tex.z-dn.net/?f=I_%7Bupper%7D%3D%20%5Cfrac%7B2%20%5Cpi%20%280.314%2A10%5E%7B-2%7D%283.30%29%7D%7B%20%284%20%5Cpi%20%2A10%5E%7B-7%7D%5Cfrac%7BT.m%7D%7BA%7D%29%281.30A%29%2815.70m%29%7D)
![I_{upper}= 2538.46 A](https://tex.z-dn.net/?f=I_%7Bupper%7D%3D%202538.46%20A)
≅ 2539 A
However; the current needed in the upper coil in each turn will be:
![I=\frac{I_{upper}}{50 turns}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BI_%7Bupper%7D%7D%7B50%20turns%7D)
![I= \frac{2539}{50}](https://tex.z-dn.net/?f=I%3D%20%5Cfrac%7B2539%7D%7B50%7D)
![I = 50.78 A](https://tex.z-dn.net/?f=I%20%3D%2050.78%20A)