A horizontal ,straight wire carying 12.0A current from west to east is in the earth's magnetic field B.At this place,B is parallel to the surface of the earth ,points to the north and its magnitude is 0.04mT.Determine the magnetic force on 1m length of the wire.If mass of this length of wire is 50g.calculate the value of current in the wire so that its weight is balanced by the magnitude force.
a) F = iLB
b) mg = iLb so i =mg/LB
To find the magnetic force on a wire, you can use the formula:
F = |B| * I * L * sin(θ)
where:
- F is the magnetic force
- |B| is the magnitude of the magnetic field
- I is the current in the wire
- L is the length of the wire
- θ is the angle between the wire and the magnetic field direction
In this case, the wire is horizontal, so the angle θ is 90 degrees. The magnitude of the magnetic field |B| is given as 0.04 mT.
Substituting the given values, we get:
F = (0.04 mT) * (12.0 A) * (1m) * sin(90°)
sin(90°) is equal to 1, so the equation simplifies to:
F = (0.04 mT) * (12.0 A) * (1m) = 0.48 mN
Therefore, the magnetic force on 1m length of the wire is 0.48 mN.
To calculate the current in the wire required to balance the weight, we can set the magnetic force equal to the weight. The weight of the wire can be calculated using the formula:
Weight = mass * g
where:
- mass is the mass of the wire (50g)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the given values, we get:
Weight = (50g) * (9.8 m/s^2)
Since the weight should be balanced by the magnetic force, we have:
Weight = Magnetic force = 0.48 mN
Substituting the known values, we can solve for the current I:
(50g) * (9.8 m/s^2) = (0.48 mN)
First, convert 50g to kg:
Weight = (0.05 kg) * (9.8 m/s^2)
Setting the two equations equal to each other:
(0.05 kg) * (9.8 m/s^2) = (0.48 mN)
We need to convert 0.48 mN to Newtons by dividing by 1000:
0.48 mN = 0.00048 N
Then solve for I:
I = (0.00048 N) / (1m) = 0.00048 A
Therefore, the value of current in the wire required to balance its weight with the magnetic force is approximately 0.00048 A.