Question 2 of 7 Attempt 3 A circular coil with 23 turns that has a radius of 24.9 cm carries a current of 2.49 A . What is the strength of the magnetic field at a point on the coil's axis that is 31.9 cm from the center of the coil

6.73 × 10⁻⁵

Explanation:

It is given that,

Number of turns, N = 23

Radius of the coil, r = 24.9 cm = 0.249 m

The distance  coil axis 31.9 cm, x = 0.319m

Current flowing through the coils, I = 2.49 A

We need to find the magnitude of the magnetic field at a location on the axis of the coils

. The magnetic field of the coils is given by :

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s