Question

A current balance is a device to measure magnetic forces. It is constructed from two parallel coils, each with an average radius of 12.5 cm. The lower coil rests on a balance; it has 20 turns and carries a constant current of 1.30 A. The upper coil, suspended 0.314 cm above the lower coil, has 50 turns and a current that can be varied. The reading of the balance changes as the magnetic force on the lower coil changes. What current is needed in the upper coil to exert a force of 3.30 N on the bottom coil

Answer 1

Answer:

$I = 50.78 A$

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:

$L__L}=N__L}(2 \pi r)$

where:

$N__L$ = 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:

$L__L}=20*(2 \pi (12.5*10^{-2}m)$

$L__L}=15.70m$

From our parameters above:

$I__L$ = 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 = $I__L}L__L}[\frac{u__0I__upper}{2 \pi d}]$

Making the varied current in the upper coil the subject of the formula; we have:

$I_{upper}= \frac{2 \pi d F}{U_oI__L}L__L}$

where : $U__0}= 4 \pi *10^{-7}\frac{T.m}{A}$

Then:

$I_{upper}= \frac{2 \pi (0.314*10^{-2}(3.30)}{ (4 \pi *10^{-7}\frac{T.m}{A})(1.30A)(15.70m)}$

$I_{upper}= 2538.46 A$

≅ 2539 A

However; the current needed in the upper coil in each turn will be:

$I=\frac{I_{upper}}{50 turns}$

$I= \frac{2539}{50}$

$I = 50.78 A$