A particular coil has 100 turns and a diameter of 6.0 m. When it's time for a measurement, a 4.5 A current is turned on. The large diameter of the coil means that the field in the water flowing directly above the center of the coil is approximately equal to the field in the center of the coil. The field is directed downward and the water is flowing east. The water is flowing above the center of the coil at 1.5 m/s .

What is the magnitude of the field at the center of the coil?

The magnetic field at the center of the coil = 5.23 * 10 ^ -5 T

Explanation:

Information from the question:

Number of turns of the coil = 100 turns

The diameter of the coil = 6 m

The radius of the coil = diameter / 2 = 3 m

The coil current = 2.5 A

Formula : The Magnetic field at the center of the coil =

                                  k * number of turns * current / 2 * radius

Therefore, The Magnetic field at the center of the coil=

                                 (4 * * 10 ^ -7 * 100 * 2.5 ) / (2 * 3)

The Magnetic field at the center of the coil = 5.23 * 10 ^ -5 T

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.