A horizontal pull A pulls two wagons over a horizontal frictionless floor, the first wagon is 500N, the second is 2000 N. The te
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Answer:
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:
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:
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 =
Making the varied current in the upper coil the subject of the formula; we have:
where :
Then:
≅ 2539 A
However; the current needed in the upper coil in each turn will be: