The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long alnico magnet is 0.10 t. to produce the same field with a solenoid of the same size, carrying a current of 2.1 a , how many turns of wire would you need?

N = 3032  turns

Explanation:

The magnetic field produced by a solenoid is described by

        B = μ₀ n I

Where is the permittivity in a vacuum with a value of 4π 10⁻⁷ N /A²,  n  is the turn density and I the current

Let's apply this equation to the problem, the turn density is the number of turns per unit length, in this case it is the same magnet length

       L = 8 cm = 0.08 m

Let's calculate

      B = μ₀ N/L   I

      N = B L / μ₀ I

      N = 0.10 0.08 / (4π 10⁻⁷  2.1)

      N = 3,032 103 turns

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s