Assume the Earth's magnetic field is 51.3 µT northward at 60.0° below the horizontal in a particular city. A tube in a neon sign stretches between two diagonally opposite corners of a shop window—which lies in a north–south vertical plane—and carries current 35.7 mA. The current enters the tube at the bottom south corner of the shop's window. It exits at the opposite corner, which is 1.39 m farther north and 0.850 m higher up. Between these two points, the glowing tube spells out DONUTS. Determine the total vector magnetic force on the tube. Hint: Two very important general statements are as follows. First, the force on the curved portion is the same in magnitude as the force on a straight wire between the same two points. In general, the magnetic force on a curved current-carrying wire in a uniform magnetic field is equal to that on a straight wire connecting the endpoints and carrying the same current. Furthermore, 1 + 2 = 0 is also a general result: the net magnetic force acting on any closed current loop in a uniform magnetic field is zero. (Take east to be the +x-axis, up to be the +y-axis, and south to be the +z-axis.)

To determine the total vector magnetic force on the neon sign tube, we need to calculate the forces on each straight segment of the tube and then combine them.

1. First, let's break down the problem into individual segments. In this case, the tube is divided into five segments: D-O, O-N, N-U, U-T, and T-S. We'll calculate the force on each segment separately.

2. Next, we need to calculate the length and direction of each segment. Since the tube is a diagonal line, we can use the Pythagorean theorem to find the length of each segment.

- The length of the DO segment is 1.39 m (given).
- The length of the ON segment is the same as the DO segment.
- The length of the NU segment is 0.850 m (given).
- The length of the UT segment is the same as the NU segment.
- The length of the TS segment can be found using the Pythagorean theorem:
- TS = sqrt(Do^2 + NU^2)

3. Now let's calculate the direction of each segment. We're given that the magnetic field is 51.3 µT northward at 60.0° below the horizontal. To determine the direction of the magnetic force, we'll need to use the right-hand rule.

- For the DO segment, the magnetic force is towards the east direction (-x-axis) using the right-hand rule.
- For the ON segment, the magnetic force is towards the east direction (-x-axis) using the right-hand rule.
- For the NU segment, the magnetic force is towards the north direction (+y-axis) using the right-hand rule.
- For the UT segment, the magnetic force is towards the north direction (+y-axis) using the right-hand rule.
- For the TS segment, we'll need to calculate the angle with respect to the horizontal. We can use the given angle of 60.0° and subtract it from 90° (since the north-south vertical plane is at 90° with respect to the horizontal), which gives us 30.0°. The magnetic force will be towards the southeast direction (between east and south) at an angle of 30.0° from the -x-axis.

4. Now we can calculate the magnetic force on each segment using the magnetic force formula:

- Magnetic Force = (Current) * (Length) * (Magnetic Field) * (sinθ)

- For the DO and ON segments, the magnetic force will be the same since they have the same length and are in the same direction.
- Magnetic Force on DO and ON segments = (35.7 mA) * (1.39 m) * (51.3 µT) * (sin(90°)) = 0 N

- For the NU and UT segments, the magnetic force will also be the same since they have the same length and are in the same direction.
- Magnetic Force on NU and UT segments = (35.7 mA) * (0.850 m) * (51.3 µT) * (sin(90°)) = 0 N

- For the TS segment, we need to calculate the component of the magnetic field that is perpendicular to the segment. This can be done using trigonometry:
- Component of Magnetic Field perpendicular to TS segment = (Magnetic Field) * (cosθ)
- cos(30.0°) = √3/2
- Component of Magnetic Field perpendicular to TS segment = (51.3 µT) * (√3/2)

- Magnetic Force on TS segment = (35.7 mA) * (TS length) * (Component of Magnetic Field perpendicular to TS segment) = (35.7 mA) * (TS length) * (51.3 µT) * (√3/2)

5. Finally, we can calculate the total vector magnetic force on the neon sign tube by summing up the magnetic forces of each segment. Since the forces on the DO, ON, NU, and UT segments are all zero, we only need to consider the magnetic force on the TS segment.

- Total Vector Magnetic Force = Magnetic Force on TS segment * Direction Unit Vector

Remember to convert between different units when necessary, such as mA to A and µT to T, to ensure consistent units in the calculations.