Answer :
Answer:
[tex]I = 50.78 A[/tex]
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:
[tex]L__L}=N__L}(2 \pi r)[/tex]
where:
[tex]N__L[/tex] = 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:
[tex]L__L}=20*(2 \pi (12.5*10^{-2}m)[/tex]
[tex]L__L}=15.70m[/tex]
From our parameters above:
[tex]I__L[/tex] = 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 = [tex]I__L}L__L}[\frac{u__0I__upper}{2 \pi d}][/tex]
Making the varied current in the upper coil the subject of the formula; we have:
[tex]I_{upper}= \frac{2 \pi d F}{U_oI__L}L__L}[/tex]
where : [tex]U__0}= 4 \pi *10^{-7}\frac{T.m}{A}[/tex]
Then:
[tex]I_{upper}= \frac{2 \pi (0.314*10^{-2}(3.30)}{ (4 \pi *10^{-7}\frac{T.m}{A})(1.30A)(15.70m)}[/tex]
[tex]I_{upper}= 2538.46 A[/tex]
≅ 2539 A
However; the current needed in the upper coil in each turn will be:
[tex]I=\frac{I_{upper}}{50 turns}[/tex]
[tex]I= \frac{2539}{50}[/tex]
[tex]I = 50.78 A[/tex]