The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 2.5 AA , how many turns of wire would you need?

Given Information:  

Current = I = 2.5 A  

Magnetic field = B = 0.10 T  

Radius = r = d/2 = 0.02/2 = 0.01 m

Length = L = 8 cm = 0.08 m

Required Information:  

Number of turns = N = ?  

Number of turns = N ≈ 2547 turns

Step-by-step explanation:  

The approximate model to find the number of turns is given by

B = μ₀nI

Where n = N/L

so

B = μ₀NI/L

N = BL/μ₀I  

Where B is the magnetic field, L is the length of the solenoid, I is the current and μ₀ is the permeability of free space

N = (0.10*0.08)/(4πx10⁻⁷*2.5)

N ≈ 2547 Turns

explanation:

as the height from which the ruler was dropped increased, the speed of the ruler at ryan's hand increased.

f=ma newton's second law 7x8=56n