The magnetic force on a straight 0.36 m segment of wire carrying a current of 4.5 A is
0.38 N.
What is the magnitude of the component
of the magnetic field that is perpendicular to
the wire?
Answer in units of T
F=ILB
B=F/IL
To calculate the magnitude of the component of the magnetic field that is perpendicular to the wire, we need to use the formula for the magnetic force on a current-carrying wire.
The formula for the magnetic force F on a current-carrying wire of length L in a magnetic field B is given by:
F = BIL sin(θ)
where:
F is the magnitude of the magnetic force,
B is the magnitude of the magnetic field,
I is the current,
L is the length of the wire segment, and
θ is the angle between the direction of the current and the magnetic field.
In this case, we are given the following information:
Length of the wire segment: L = 0.36 m
Current: I = 4.5 A
Magnetic force: F = 0.38 N
We need to rearrange the formula to solve for B:
B = F / (IL sin(θ))
Since the problem states that we need to find the magnitude of the component of the magnetic field perpendicular to the wire, we can assume that θ = 90 degrees.
Now we can plug in the values into the formula:
B = 0.38 N / (4.5 A * 0.36 m * sin(90))
Note: sin(90) equals 1, so the term sin(θ) simplifies to 1.
B = 0.38 N / (4.5 A * 0.36 m * 1)
B = 0.38 N / 1.62 Am
B ≈ 0.235 T
Therefore, the magnitude of the component of the magnetic field that is perpendicular to the wire is approximately 0.235 T.