Finding the Minimum Current for Magnetic Force to Equal Wire Weight

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To determine the minimum current required for the magnetic force on a wire to equal its weight, the equation F = BIl sin θ is used, where B is the magnetic field strength, I is the current, and l is the length of the wire. The wire has a mass of 0.18g per meter and carries a current of 1A in a magnetic field of 0.5T directed horizontally upwards. The weight of the wire (mg) is calculated as 0.18E-3 kg multiplied by 10 m/s², resulting in a force of 1.8E-3 N. By equating the magnetic force to the weight, the equation simplifies to 0.5I = 1.8E-3, leading to a minimum current of 0.0036 A. This calculation confirms that the current must be at least 0.0036 A for the magnetic force to balance the wire's weight.
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Homework Statement



A wire extends horizontally from east to west. The mass of its meter is (0.18)g and it carries a current of (1)A. If the wire is in a magnetic filed (0.5)T that heads horizontally upwards, determine the least current that makes the magnetic force equal the wire weight

Homework Equations



F = BIl sin θ

The Attempt at a Solution



at first i don't understand what does "heads horizontally upwards" means so this is where i end up:

mg = BIl sin θ
(0.18E-3) x 10 = 0.5x1x1 sin θ
θ = sin^-1(1.8E-3/0.5)
θ = 0.2
 
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I think horizontally upwards means perpendicular to the wire(direction of the current)
 
You mean..

BIl = mg
0.5 I = 1.8E-3
I = 0.0036 A ?
 

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