A straight, vertical wire carries a current of 1.27 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has a magnitude of B = 0.592 T and is horizontal.

PART A: What is the magnitude of the magnetic force on a section of the wire with a length of 1.00 cm that is in this uniform magnetic field if the magnetic field direction is east?
PART C: What is the magnitude of the magnetic force on a section of the wire with a length of 1.00 cm that is in this uniform magnetic field, if the magnetic field direction is south?
PART E: What is the magnitude of the magnetic force on a section of the wire with a length of 1.00 cm that is in this uniform magnetic field, if the magnetic field direction is 32.0 degrees south of west.
PART F: What angle will the magnetic force on this segment of wire make relative to the north, if the magnetic field direction is 32.0 degrees south of west?

a) F = 7.52 10⁻³ N, north c) 7.52 10⁻³ N, est, e)    F = 3.98 10⁻³ N, f) θ = 58º north west

Explanation:

The magnetic force is given by the expression

          F = q v x B = i L x B

Where the bold indicate vectors, L is the wire length and B  magnetic field

The direction of the field can be found with the rule of the right hand, the extended fingers point in the direction of the magnetic field, the thumb in the direction of the current and the palm of the hand is the direction of the magnetic force.

Let's apply to our case

The field is horizontal and the current is vertical (z axis), so it always has 90 degrees between them

a) the field goes east and the current is vertical, so the angle is 90

            F = i L B

            F = 1.27 0.01 0.592

            F = 7.52 10⁻³ N

The direction is east field, vertical current, the force is north

c) the field points south

       F = i L B

       F = 7.52 10⁻³ N

Field to the south, vertical current (z), force (palm) points east

e) field 32 south west

     F = i L B sin 32

     F = 3.98 10⁻³ N

f) The current thumb) points to the z-axis, those of two extended in the direction of the field and the palm is the direction of the force, bone perpendicular to the direction of 32 south west

         θ = 90 - 32

         θ = 58º north west

v 154.9= 0.9 length length s little over 154.9cm