A copper wire 0.40m long carries a current of 6.0A and weighs 0.35N. What strenght magnetic field is required to balance the weight of the wire?
Make the vertical force BIL equal the weight, 0.35 N. Solve for B, in Tesla.
L is the wire length
To find the strength of the magnetic field required to balance the weight of the wire, we can use the equation:
F = BIL
where:
F is the weight of the wire (0.35N)
B is the magnetic field strength
I is the current flowing through the wire (6.0A)
L is the length of the wire (0.40m)
Since we want to balance the weight of the wire, the magnetic force (BIL) should be equal to the weight of the wire (F).
Therefore, we can rearrange the equation to solve for B:
B = F / (IL)
Substituting the given values, we have:
B = 0.35N / (6.0A * 0.40m)
Calculating this, we find:
B = 0.35N / 2.4A·m
B ≈ 0.146 T (rounded to three decimal places)
Therefore, a magnetic field strength of approximately 0.146 Tesla is required to balance the weight of the wire.